CHAPTER SIX

Coordinate Geometry
Coordinate geometry is geometry done in the number plane, using algebra.
• Points are represented by pairs of numbers, and lines by linear equations.
• Circles, parabolas and other curves are represented by non-linear equations.
Points, lines and intervals are the main topics of this chapter.

6 A Lengths and Midpoints of Intervals
An interval is completely determined by its two endpoints. There are simple formulae for the length of an interval and for the midpoint of an interval.

The Distance Formula: The formula for the length of an inter-

y

val P Q is just Pythagoras’ theorem in different notation.
Let P (x1 , y1 ) and Q(x2 , y2 ) be two points in the plane.
Construct the right-angled triangle P QA, where A(x2 , y1 ) lies level with P and vertically above or below Q.
Then P A = |x2 − x1 | and QA = |y2 − y1 |, and so by Pythagoras’ theorem in P QA,

Q(x2,y2)

y2 y1 P(x1,y1) x1 A x2 x

P Q2 = (x2 − x1 )2 + (y2 − y1 )2 .

DISTANCE FORMULA: Let P (x1 , y1 ) and Q(x2 , y2 ) be two points in the plane. Then
1

P Q2 = (x2 − x1 )2 + (y2 − y1 )2 .
• First find the square P Q2 of the distance.
• Then take the square root to find the distance P Q.

WORKED EXERCISE:

Find the lengths of the sides AB and AC of the triangle with vertices A(1, −2),
B(−4, 2), and C(5, −7), and hence show that ABC is isosceles.

SOLUTION:
First, AB 2 = (x2 − x1 )2 + (y2 − y1 )2
2

so

= − 4 − 1 + 2 − (−2)
= (−5)2 + 42
= 41,
√
AB = 41 .

Secondly, AC 2 = (x2 − x1 )2 + (y2 − y1 )2
2

2

so

= 5 − 1 + − 7 − (−2)
= 42 + (−5)2
= 41,
√
AC = 41 .

2

Since the two sides AB and AC are equal, the triangle is isosceles.

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The Midpoint Formula: The midpoint of an interval is found by taking the averages of the coordinates of the two points. Congruence is the basis of the proof below.
Let P (x1 , y1 ) and Q(x2 , y2 ) be two points in the plane, and let M (x, y) be the midpoint of P Q. y Construct S(x, y1 ) and T (x2 , y), as shown.
Q(x2,y2)
Then
P M S ≡ M QT (AAS)
M(x,y)
and so
P S = M T (matching sides of congruent triangles).
T
That is, x − x1 = x2 − x
P(x1,y1)
S
2x = x1 + x2 x1 + x2 x= , which is the average of x1 and x2 . x1 x x2 2
The calculation of the y-coordinate y is similar.

x

MIDPOINT FORMULA:
Let M (x, y) be the midpoint of the interval joining P (x1 , y1 ) and Q(x2 , y2 ). Then
2
x=

x1 + x2
2

and

y=

y1 + y2
,
2

(Take the average of the coordinates.)

WORKED EXERCISE:

The interval joining the points A(3, −1) and B(−7, 5) is a diameter of a circle.
(a) Find the centre M of the circle.
(b) Find the radius of the circle.

SOLUTION:
(a) The centre of the circle is the midpoint M (x, y) of the interval AB. x1 + x2 y1 + y2
Using the midpoint formula, x = and y =
2
2
3−7
−1 + 5
=
=
2
2
= −2,
= 2, so the centre is M (−2, 2).
B

(b) Using the distance formula, AM 2 = (x2 − x1 )2 + (y2 − y1 )2
= −2−3
= 34,
√
AM = 34 .
√
Hence the circle has radius 34 .

2

+ 2 − (−1)

y

2

5
3

M
-7

-1

A x

Testing for Special Quadrilaterals: Euclidean geometry will be reviewed in the Year 12 volume, but many questions in this chapter ask for proofs that a quadrilateral is of a particular type. The most obvious way is to test the definition itself.

3

DEFINITIONS OF THE SPECIAL QUADRILATERALS:
• A trapezium is a quadrilateral with one pair of opposite sides parallel.
• A parallelogram is a quadrilateral with both pairs of opposite sides parallel.
• A rhombus is a parallelogram with a pair of adjacent sides equal.
• A rectangle is a parallelogram with one angle a right angle.
• A square is both a rectangle and a rhombus.

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CHAPTER 6: Coordinate Geometry

6A Lengths and Midpoints of Intervals

145

There are, however, several further standard tests that the exercises assume.
(Tests involving angles are omitted here, being irrelevant in this chapter.)

A QUADRILATERAL IS A PARALLELOGRAM:
• if the opposite sides are equal, or
• if one pair of opposite sides are equal and parallel, or
• if the diagonals bisect each other.
4

A QUADRILATERAL IS A RHOMBUS:
• if all sides are equal, or
• if the diagonals bisect each other at right angles.
A QUADRILATERAL IS A RECTANGLE:
• if the diagonals are equal and bisect each other.

Exercise 6A
Note:

Diagrams should be drawn wherever possible. y1 + y2 x1 + x2 and y =
.
2
2
(d) A(−3, 6) and B(3, 1)
(e) A(0, −8) and B(−11, −12)
(f) A(4, −7) and B(4, 7)

1. Find the midpoint of each interval AB. Use x =
(a) A(3, 5) and B(1, 9)
(b) A(4, 8) and B(6, 4)
(c) A(−4, 7) and B(8, −11)

2. Find the length of each interval. Use AB 2 = (x2 − x1 )2 + (y2 − y1 )2 , then find AB.
(a) A(1, 4), B(5, 1)
(d) A(3, 6), B(5, 4)
(b) A(−2, 7), B(3, −5)
(e) A(−4, −1), B(4, 3)
(c) A(−5, −2), B(3, 4)
(f) A(5, −12), B(0, 0)
3. (a) Find the midpoint M of the interval joining P (−2, 1) and Q(4, 9).
(b) Find the lengths P M and M Q, and verify that P M = M Q.
4. (a) Find the length of each side of the triangle formed by P (0, 3), Q(1, 7) and R(5, 8).
(b) Hence show that P QR is isosceles.
5. (a) Find the length of each side of ABC, where A = (0, 5), B = (3, −2) and C = (−3, 4).
(b) Find the midpoint of each side of this triangle ABC.
DEVELOPMENT

6. (a)
(b)
(c)
(d)
(e)

A circle with centre O(0, 0) passes through A(5, 12). What is its radius?
A circle with centre B(4, 5) passes through the origin. What is its radius?
Find the centre of the circle with diameter CD, where C = (2, 1) and D = (8, −7).
Find the radius of the circle with diameter CD in part (c) above.
Show that E(−12, −5) lies on the circle with centre the origin and radius 13.

7. The interval joining A(2, −5) and E(−6, −1) is divided into four equal subintervals by the three points B, C and D.
(a) Find the coordinates of C by taking the midpoint of AE.
(b) Find the coordinates of B and D by taking the midpoints of AC and CE.

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CHAPTER 6: Coordinate Geometry

8. (a)
(b)
(c)
(d)

CAMBRIDGE MATHEMATICS 2 UNIT YEAR 11

r

Find the midpoint of the interval joining A(4, 9) and C(−2, 3).
Find the midpoint of the interval joining B(0, 4) and D(2, 8).
What can you conclude about the diagonals of the quadrilateral ABCD?
What sort of quadrilateral is ABCD? [Hint: See Box 4 above.]

9. The points A(3, 1), B(10, 2), C(5, 7) and D(−2, 6) are the vertices of a quadrilateral.
(a) Find the lengths of all four sides.
(b) What sort of quadrilateral is ABCD? [Hint: See Box 4 above.]
10. (a) Find the side lengths of the triangle with vertices X(0, −4), Y (4, 2) and Z(−2, 6).
(b) Show that XY Z is a right-angled isosceles triangle by showing that its side lengths satisfy Pythagoras’ theorem.
(c) Hence find the area of XY Z.
11. The quadrilateral ABCD has vertices at the points A(1, 0), B(3, 1), C(4, 3) and D(2, 2).
[Hint: You should look at Boxes 3 and 4 in the notes above to answer this question.]
(a) Show that the intervals AC and BD bisect each other, by finding the midpoint of each and showing that these midpoints coincide.
(b) What can you conclude from part (a) about what type of quadrilateral ABCD is?
(c) Show that AB = AD. What can you now conclude about the quadrilateral ABCD?
√
√
12. (a) Find the distance of each point A(1, 4), B(2, 13 ), C(3, 2 2 ) and D(4, 1) from the origin O. Hence explain why the four points lie on a circle with centre the origin.
(b) What are the radius, diameter, circumference and area of this circle?
13. The point M (3, 7) is the midpoint of the interval joining A(1, 12) and B(x2 , y2 ). Find the coordinates x2 and y2 of B by substituting into the formulae y1 + y2 x1 + x2 and y=
.
x=
2
2
14. Solve each of these problems using the same methods as in the previous question.
(a) If A(−1, 2) is the midpoint of S(x, y) and T (3, 6), find the coordinates of S.
(b) The midpoint of the interval P Q is M (2, −7). Find the coordinates of P if:
(i) Q = (0, 0)
(ii) Q = (5, 3)
(iii) Q = (−3, −7)
(c) Find B, if AB is a diameter of a circle with centre Q(4, 5) and A = (8, 3).
(d) Given that P (4, 7) is one vertex of the square P QRS, and that the centre of the square is M (8, −1), find the coordinates of the opposite vertex R.
15. (a) Write down any two points A and B whose midpoint is M (4, 6).
(b) Write down any two points C and D that are 10 units apart.
CHALLENGE

16. Each set of three points given below forms a triangle of one of these types:
A. isosceles,

B. equilateral,

C. right-angled,

D. none of these.

Find the side lengths of each triangle below and hence determine its type.
√
(c) D(1, 1), E(2, −2), F (−3, 0)
(a) A(−1, 0), B(1, 0), C(0, 3)
(d) X(−3, −1), Y (0, 0), Z(−2, 2)
(b) P (−1, 1), Q(0, −1), R(3, 3)
17. As was discussed in Section 3F, the circle with centre (h, k) and radius r has equation
(x − h)2 + (y − k)2 = r2 . By identifying the centre and radius, find the equations of:
(a) the circle with centre (5, −2) and passing through (−1, 1),
(b) the circle with K(5, 7) and L(−9, −3) as endpoints of a diameter.

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CHAPTER 6: Coordinate Geometry

6B Gradients of Intervals and Lines

147

6 B Gradients of Intervals and Lines
Gradient is the key idea that will be used in the next section to bring lines and their equations into coordinate geometry.

y
Q(x2,y2)

y2

The Gradient of an Interval and the Gradient of a Line:

rise

Let P (x1 , y1 ) and Q(x2 , y2 ) be two points in the plane. y1 The gradient of the interval P Q is a measure of its steepness,
P(x1,y1)
as someone walks along the interval from P to Q. x1 x2 run Their rise is the vertical difference y2 − y1 , and their run is the horizontal difference x2 − x1 .
The gradient of the interval P Q is defined to be the ratio of the rise and the run: rise gradient of P Q = run y2 − y1
.
= x2 − x1

x

THE GRADIENT OF AN INTERVAL:
Let P (x1 , y1 ) and Q(x2 , y2 ) be any two points. Then rise run y2 − y1
.
gradient of P Q = x2 − x1 gradient of P Q =

OR

5

• Horizontal intervals have gradient zero, because the rise is zero.
• Vertical intervals don’t have a gradient, because the run is always zero and so the fraction is undefined.

THE GRADIENT OF A LINE:
The gradient of a line is then determined by taking any two points A and B on the line and finding the gradient of the interval AB.

Positive and Negative Gradients: If the rise and the run have the same sign, then the gradient is positive, as in the first diagram below. In this case the interval slopes upwards as one moves from left to right.
If the rise and run have opposite signs, then the gradient is negative, as in the second diagram. The interval slopes downwards as one moves from left to right. y y
Q

P

P

Q x x

If the points P and Q are interchanged, then the rise and the run both change signs, but the gradient remains the same.

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WORKED EXERCISE:
Find the gradients of the sides of

SOLUTION:

y2 − y1 x2 − x1
−2 − 5
=
5−2
−7
=
3
=−7
3

gradient of XY =

XYZ, where X = (2, 5), Y = (5, −2) and Z = (−3, 4). y2 − y1 x2 − x1
4 − (−2)
=
−3 − 5
4+2
=
−8
=−3
4

gradient of Y Z =

A Condition for Two Lines to be Parallel: Any two vertical lines are

y2 − y1 x2 − x1
5−4
=
2 − (−3)
1
=
2+3
=1
5

gradient of ZX =

y

parallel. Otherwise, the condition for lines to be parallel is:

6

r

Q

l1

B

l2

PARALLEL LINES: Two lines 1 and 2 with gradients m1 and m2 are parallel if and only if m1 = m2 .

This can easily be seen by looking at the congruent triangles in the diagram to the right.

P

R

A

C

x

WORKED EXERCISE:

Given the four points A(3, 6), B(7, −2), C(4, −5) and D(−1, 5), show that the quadrilateral ABCD is a trapezium with AB CD.

SOLUTION:

−2 − 6
7−3
−8
=
4
= −2,

gradient of AB =

Hence AB

5 − (−5)
−1 − 4
10
=
−5
= −2.

gradient of CD =

CD because their gradients are equal, so ABCD is a trapezium.

Testing for Collinear Points: Three points are called collinear if they all lie on one line.

7

TESTING FOR COLLINEAR POINTS: To test whether three given points A, B and C are collinear, find the gradients of AB and BC.
If these gradients are equal, then the three points must be collinear, because then
AB and BC are parallel lines passing through a common point B.

WORKED EXERCISE:

Test whether the three points A(−2, 5), B(1, 3) and C(7, −1) are collinear.

SOLUTION:

3−5
1+2
= −2 ,
3

gradient of AB =

−1 − 3
7−1
= −2 .
3

gradient of BC =

Since the gradients are equal, the points A, B and C are collinear.

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CHAPTER 6: Coordinate Geometry

6B Gradients of Intervals and Lines

149

Gradient and the Angle of Inclination: The angle of inclination of a line is the angle between the upward direction of the line and the positive direction of the x-axis. y y

P
P

O

α

α
M x

M

x

O

The two diagrams above show that lines with positive gradients have acute angles of inclination, and lines with negative gradients have obtuse angles of inclination.
They also illustrate the trigonometric relationship between the gradient and the angle of inclination α.

ANGLE OF INCLINATION: Let the line
8

gradient of

have angle of inclination α. Then

= tan α.

Proof:
A. When α is acute, as in the first diagram, then the rise M P and the run OM are just the opposite and adjacent sides of the triangle P OM , so tan α =

MP
= gradient of OP.
OM

B. When α is obtuse, as in the second diagram, then P OM = 180◦ − α, so tan α = − tan P OM = −

MP
= gradient of OP.
OM

WORKED EXERCISE:
(a) Given the points A(−3, 5), B(−6, 0) and O(0, 0), find the angles of inclination of the intervals AB and AO.
(b) What sort of triangle is ABO?

SOLUTION:
(a) First,

0−5
−6 + 3
= 5,
3

gradient of AB =

and using a calculator to solve tan α = 5 ,
3

. angle of inclination of AB = 59◦ .
.
0−5
Secondly,
gradient of AO =
0+3
= −5 ,
3

and using a calculator to solve tan α = − 5 ,
3

A

B
−6

−3

y

O

5

x

. angle of inclination of AO = 121◦ .
.

(b) Hence
Thus the base angles of

AOB = 59◦ (straight angle).
AOB are equal, and so the triangle is isosceles.

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A Condition for Lines to be Perpendicular: A vertical and a horizontal line are perpendicular. Otherwise, the condition is: y PERPENDICULAR LINES: Lines 1 and 2 with gradients m1 and m2 are perpendicular if and only if

that is,

m2 = −

P

l2

m1 × m2 = −1

9

B

A

1
.
m1

l1

O

Q x

Proof:
Shift each line sideways, without rotating it, so that it passes through the origin.
One line must have positive gradient and the other negative gradient, otherwise one of the angles between them would be acute.
So let 1 be a line with positive gradient through the origin, and let 2 be a line with negative gradient through the origin.
Construct the two triangles P OQ and AOB as shown in the diagram above, with the run OQ of 1 equal to the rise OB of 2 .
QP
OB
QP
Then m1 × m2 =
× −
=−
, since OQ = OB.
OQ
AB
AB
A. If the lines are perpendicular, then AOB = P OQ. (adjacent angles at O)
Hence
AOB ≡ P OQ (AAS) so QP = AB (matching sides of congruent triangles) and so m1 × m2 = −1.
B. Conversely, if m1 × m2 = −1, then QP = AB.
Hence AOB ≡ P OQ (SAS)
AOB = P OQ, (matching angles of congruent triangles) so and so 1 and 2 are perpendicular.

WORKED EXERCISE:
What is the gradient of a line perpendicular to a line with gradient 2 ?
3

SOLUTION:
Perpendicular gradient = − 3 . (Take the opposite of the reciprocal of 2 .)
2
3

WORKED EXERCISE:
Show that the diagonals of the quadrilateral ABCD are perpendicular, where the vertices are A(3, 7), B(−1, 6), C(−2, −3) and D(11, 0).

SOLUTION:

y2 − y1 x2 − x1
−3 − 7
=
−2 − 3
= 2,

Gradient of AC =

y2 − y1 x2 − x1
0−6
=
11 + 1
= −1 .
2

gradient of BD =

Hence AC ⊥ BD, because the product of the gradients of AC and BD is −1.

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CHAPTER 6: Coordinate Geometry

6B Gradients of Intervals and Lines

151

WORKED EXERCISE:
The interval joining the points C(−6, 0) and D(−1, a) is perpendicular to a line with gradient 10. Find the value of a.

SOLUTION: a−0 −1 + 6 a = .
5
Since the interval CD is perpendicular to a line with gradient 10, a 10
×
= −1
(The product of the gradients is −1.)
5
1 a × 2 = −1
The interval CD has gradient =

÷2

a = −1 .
2

Exercise 6B
Note:

Diagrams should be drawn wherever possible.

1. (a) Write down the gradient of a line parallel to a line with gradient:
(i) 2
(ii) −1
(iii) 3
4
(b) Find the gradient of a line perpendicular to a line with gradient:
(i) 2
(ii) −1
(iii) 3
4

(iv) −1 1
2
(iv) −1 1
2

y2 − y1 to find the gradient of each interval AB below. Then x2 − x1 find the gradient of a line perpendicular to it.
(a) A(1, 4), B(5, 0)
(c) A(−5, −2), B(3, 2)
(e) A(−1, −2), B(1, 4)
(b) A(−2, −7), B(3, 3)
(d) A(3, 6), B(5, 5)
(f) A(−5, 7), B(15, −7)

2. Use the formula gradient =

3. The points A(2, 5), B(4, 11), C(12, 15) and D(10, 9) form a quadrilateral.
(a) Find the gradients of AB and DC, and hence show that AB DC.
(b) Find the gradients of BC and AD, and hence show that BC AD.
(c) What type of quadrilateral is ABCD? [Hint: Look at the definitions in Box 3 above.]
4. (a) Show that A(−2, −6), B(0, −5), C(10, −7) and D(8, −8) form a parallelogram.
(b) Show that A(6, 9), B(11, 3), C(0, 0) and D(−5, 6) form a parallelogram.
(c) Show that A(2, 5), B(3, 7), C(−4, −1) and D(−5, 2) do not form a parallelogram.
5. Use the formula gradient = tan α to find the gradient, correct to two decimal places where necessary, of a line with angle of inclination:
◦
(a) 15◦
(b) 135◦
(d) 72◦
(c) 22 1
2
6. Use the formula gradient = tan α to find the angle of inclination, correct to the nearest degree where necessary, of a line with gradient:
√
1
(a) 1
(b) − 3
(c) 4
(d) √
3

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CAMBRIDGE MATHEMATICS 2 UNIT YEAR 11

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DEVELOPMENT

7. The quadrilateral ABCD has vertices A(−1, 1), B(3, −1), C(5, 3) and D(1, 5). Use the definitions of the special quadrilaterals in Box 3 above to answer these questions.
(a) Show that the opposite sides are parallel, and hence that ABCD is a parallelogram.
(b) Show that AB ⊥ BC, and hence that ABCD is a rectangle.
(c) Show that AB = BC, and hence that ABCD is a square.
8. Use gradients to show that each quadrilateral ABCD below is a parallelogram. Then use the definitions in Box 3 of the notes to show that it is:
(a) a rhombus, for the vertices A(2, 1), B(−1, 3), C(1, 0) and D(4, −2),
(b) a rectangle, for the vertices A(4, 0), B(−2, 3), C(−3, 1) and D(3, −2),
(c) a square, for the vertices A(3, 3), B(−1, 2), C(0, −2) and D(4, −1).
9. A quadrilateral has vertices W (2, 3), X(−7, 5), Y (−1, −3) and Z(5, −1).
(a) Show that WZ is parallel to XY , but that WZ and XY have different lengths.
(b) What type of quadrilateral is WXY Z? [Hint: Look at Boxes 3 and 4 above.]
(c) Show that the diagonals W Y and XZ are perpendicular.
10. Find the gradients of P Q and QR, and hence determine whether P , Q and R are collinear.
(a) P (−2, 7), Q(1, 1), R(4, −6)
(b) P (−5, −4), Q(−2, −2), R(1, 0)
11. Show that the four points A(2, 5), B(5, 6), C(11, 8) and D(−16, −1) are collinear.
12. The triangle ABC has vertices A(−1, 0), B(3, 2) and C(4, 0). Calculate the gradient of each side and hence show that ABC is a right-angled triangle.
13. Similarly, show that each triangle below is right-angled. Then find the lengths of the sides enclosing the right angle, and calculate the area of each triangle.
(a) P (2, −1), Q(3, 3), R(−1, 4)
(b) X(−1, −3), Y (2, 4), Z(−3, 2)
14. (a) Write down two points A and B for which the interval AB has gradient 3.
(b) Write down two points A and B for which the interval AB is vertical.
(c) Write down two points A and B for which AB has gradient 2 and midpoint M (4, 6).
15. The interval P Q has gradient −3. A second line passes through A(−2, 4) and B(1, k).
(a) Find k if AB is parallel to P Q.
(b) Find k if AB is perpendicular to P Q.
16. Find the points A and B where each line below meets the x-axis and y-axis respectively.
Hence find the gradient of AB and its angle of inclination α (correct to the nearest degree).
(e) 4x − 5y − 20 = 0
(c) 3x + 4y + 12 = 0
(a) y = 3x + 6 x y x y
(f)
− =1
+ =1
(d)
(b) y = − 1 x + 1
2
3 2
2 5
17. The quadrilateral ABCD has vertices A(1, −4), B(3, 2), C(−5, 6) and D(−1, −2).
(a) Find the midpoints P of AB, Q of BC, R of CD, and S of DA.
(b) Prove that P QRS is a parallelogram by showing that P Q RS and P S QR.
18. The points A(1, 4), B(5, 0) and C(9, 8) form the vertices of a triangle.
(a) Find the coordinates of the midpoints P and Q of AB and AC respectively.
(b) Show that P Q is parallel to BC and half its length.
19. (a) Show that the points A(−5, 0), B(5, 0) and C(3, 4) all lie on the circle x2 + y 2 = 25.
(b) Explain why AB is a diameter of the circle.
(c) Show that AC ⊥ BC.

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CHALLENGE

20. Find the gradient, correct to two decimal places where appropriate, of a line sloping upwards, if its acute angle with the y-axis is:
◦
(a) 15◦
(b) 45◦
(d) 72◦
(c) 22 1
2
21. Given the points X(−1, 0), Y (1, a) and Z(a, 2), find a if YXZ = 90◦ .
22. For the four points P (k, 1), Q(−2, −3), R(2, 3) and S(1, k), it is known that P Q is parallel to RS. Find the possible values of k.

6 C Equations of Lines
In coordinate geometry, a line is represented by a linear equation in x and y.
This section and the next develop various useful forms for the equation of a line.

Horizontal and Vertical Lines: All the points on a vertical line must have the same x-coordinate, but the y-coordinate can take any value.

10

VERTICAL LINES: The vertical line through the point P (a, b) has equation x = a.

All the points on a horizontal line must have the same y-coordinate, but the x-coordinate can take any value.

HORIZONTAL LINES: The horizontal line through the point P (a, b) has equation
11

y = b. y y

x=a

y=b

P(a,b)

a

b

P(a,b)

x

x

Gradient–Intercept Form: There is a simple form of the equation of a line whose gradient and y-intercept are known.
Let have gradient m and y-intercept b, passing through the point B(0, b).
Let Q(x, y) be any other point in the plane. y Then the condition for Q to lie on the line is gradient of BQ = m,
B(0,b)
y−b that is,
=m
(This is the formula for gradient.) x−0 y − b = mx. y = mx + b.

Q(x,y)

x

GRADIENT–INTERCEPT FORM: The line with gradient m and y-intercept b is
12

y = mx + b.

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WORKED EXERCISE:
(a) Write down the gradient and the y-intercept of the line : y = 3x − 2.
(b) Find the equation of the line through B(0, 5) parallel to .
(c) Find the equation of the line through B(0, 5) perpendicular to .

SOLUTION:
(a) The line

: y = 3x − 2 has gradient 3 and y-intercept −2.

(b) The line through B(0, 5) parallel to so its equation is y = 3x + 5.

has gradient 3 and y-intercept 5,

(c) The line through B(0, 5) perpendicular to so its equation is y = − 1 x + 5.
3

has gradient − 1 and y-intercept 5,
3

General Form: It is often useful to write the equation of a line so that all the terms are on the LHS and only zero is on the RHS. This is called general form.

GENERAL FORM: The equation of a line is said to be in general form if it is ax + by + c = 0,

13

where a, b and c are constants.

• When an equation is given in general form, it should usually be simplified by multiplying out all fractions and dividing through by all common factors.

WORKED EXERCISE:
(a) Put the equation of the line 3x + 4y + 5 = 0 in gradient–intercept form.
(b) Hence write down the gradient and y-intercept of the line 3x + 4y + 5 = 0.

SOLUTION:
(a) Solving the equation for y, 4y = −3x − 5
÷4

y = − 3 x − 5 , which is gradient–intercept form.
4
4

(b) Hence the line has gradient − 3 and y-intercept −1 1 .
4
4

WORKED EXERCISE:

Find, in general form, the equation of the line passing through B(0, −2) and:
(a) perpendicular to a line with gradient 2 ,
3
(b) having angle of inclination 60◦ .

SOLUTION:
(a) The line through B perpendicular to has gradient − 3 and y-intercept −2,
2
(This is gradient–intercept form.) so its equation is y = −3 x − 2
2
×2

2y = −3x − 4
3x + 2y + 4 = 0.

√
(b) The line through B with angle of inclination 60◦ has gradient tan 60◦ = 3,
√
so its equation is y =x 3−2
(This is gradient–intercept form.)
√
x 3 − y − 2 = 0.

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CHAPTER 6: Coordinate Geometry

6C Equations of Lines

155

Exercise 6C
1. Determine, by substitution, whether the point A(3, −2) lies on the line:
(a) y = 4x − 10
(b) 8x + 10y − 4 = 0
(c) x = 3
2. Write down the coordinates of any three points on the line x + 3y = 24.
3. Write down the equations of the vertical and horizontal lines through:
(a) (1, 2)
(b) (−1, 1)
(c) (0, −4)
(d) (5, 0)
4. Write down the gradient and y-intercept of each line.
(c) y = 2 − x
(a) y = 4x − 2
(b) y = 1 x − 3
5

(e) (−2, −3)

(d) y = − 5 x
7

5. Use the formula y = mx + b to write down the equation of the line with gradient −3 and:
(a) y-intercept 5,
(b) y-intercept − 2 ,
(c) y-intercept 17 1 , (d) y-intercept 0.
3
2
6. Use the formula y = mx+b to write down the equation of the line with y-intercept −4 and:
(a) gradient 5,
(b) gradient − 2 ,
(c) gradient 17 1 ,
(d) gradient 0.
3
2
7. Use the formula y = mx + b to write down the equation of the line:
(a) with gradient 1 and y-intercept 3,
(c) with gradient 1 and y-intercept −1,
5
(b) with gradient −2 and y-intercept 5,
(d) with gradient − 1 and y-intercept 3.
2
8. Solve each equation for y and hence write down its gradient m and y-intercept b.
(a) x − y + 3 = 0
(c) 2x − y = 5
(e) 3x + 4y = 5
(b) y + x − 2 = 0
(d) x − 3y = 0
(f) 2y − 3x = −4
9. Write down the gradient m of each line. Then use the formula gradient = tan α to find its angle of inclination α, correct to the nearest minute where appropriate.
(a) y = x + 3
(b) y = −x − 16
(c) y = 2x
(d) y = − 3 x
4
DEVELOPMENT

10. Substitute y = 0 and x = 0 into the equation of each line below to find the points A and B where the line crosses the x-axis and y-axis respectively. Hence sketch the line.
(a) 5x + 3y − 15 = 0
(b) 2x − y + 6 = 0
(c) 3x − 5y + 12 = 0
11. Find the gradient of the line through each pair of given points. Then find its equation, using gradient–intercept form. Give your final answer in general form.
(c) (−9, −1), (0, −4)
(d) (2, 6), (0, 11)
(a) (0, 4), (2, 8)
(b) (0, 0), (1, −1)
12. Find the gradient of each line below. Hence find, in gradient–intercept form, the equation of a line passing through A(0, 3) and: (i) parallel to it, (ii) perpendicular to it.
(a) 2x + y + 3 = 0
(b) 5x − 2y − 1 = 0
(c) 3x + 4y − 5 = 0
13. Find the gradients of the four lines in each part. Hence state what sort of special quadrilateral they enclose.
(a) 3x + y + 7 = 0, x − 2y − 1 = 0, 3x + y + 11 = 0, x − 2y + 12 = 0
(b) 4x − 3y + 10 = 0, 3x + 4y + 7 = 0, 4x − 3y − 7 = 0, 3x + 4y + 1 = 0
14. Find the gradients of the three lines 5x − 7y + 5 = 0, 2x − 5y + 7 = 0 and 7x + 5y + 2 = 0.
Hence show that they enclose a right-angled triangle.
15. Draw a sketch of, then find the equations of the sides of:
(a) the rectangle with vertices P (3, −7), Q(0, −7), R(0, −2) and S(3, −2),
(b) the triangle with vertices F (3, 0), G(−6, 0) and H(0, 12).

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16. In each part below, the angle of inclination α and the y-intercept A of a line are given.
Use the formula gradient = tan α to find the gradient of each line, then find its equation in general form.
(a) α = 45◦ , A = (0, 3)
(c) α = 30◦ , A = (0, −2)
(b) α = 60◦ , A = (0, −1)
(d) α = 135◦ , A = (0, 1)
CHALLENGE

17. A triangle is formed by the x-axis and the lines 5y = 9x and 5y + 9x = 45.
(a) Find (correct to the nearest degree) the angles of inclination of the two lines.
(b) What sort of triangle has been formed?
18. Consider the two lines 1 : 3x − y + 4 = 0 and 2 : x + ky + = 0. Find the value of k if:
(a) 1 is parallel to 2 ,
(b) 1 is perpendicular to 2 .

6 D Further Equations of Lines
This section introduces another standard form of the equation of a line, called point–gradient form. It also deals with lines through two given points, and the point of intersection of two lines.

Point–Gradient Form: Point–gradient form gives the equation of a line with gradient m passing through a particular point P (x1 , y1 ).

y

Let Q(x, y) be any other point in the plane.
Then the condition for Q to lie on the line is gradient of P Q = m, y − y1 that is,
= m (This is the formula for gradient.) x − x1 y − y1 = m(x − x1 ).

Q(x,y)

P(x1,y1) x POINT–GRADIENT FORM: The line with gradient m through the point (x1 , y1 ) is
14

y − y1 = m(x − x1 ).

WORKED EXERCISE:
(a) Find the equation of the line through (−2, −5) and parallel to y = 3x + 2.
(b) Express the answer in gradient–intercept form, and hence find its y-intercept.

SOLUTION:
(a) The line y = 3x + 2 has gradient 3.
Hence the required line is y − y1 = m(x − x1 ) y + 5 = 3(x + 2) y + 5 = 3x + 6 y = 3x + 1.

(This is point–gradient form.)

(This is gradient–intercept form.)

(b) Hence the new line has y-intercept 1.

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CHAPTER 6: Coordinate Geometry

6D Further Equations of Lines

157

The Line through Two Given Points: Given two distinct points, there is just one line passing through them both. Its equation is best found by a two-step approach.

THE LINE THROUGH TWO GIVEN POINTS:
15

• First find the gradient of the line, using the gradient formula.
• Then find the equation of the line, using point–gradient form.

WORKED EXERCISE:

Find the equation of the line passing through A(1, 5) and B(4, −1).

SOLUTION:
First, using the gradient formula,
−1 − 5 gradient of AB =
4−1
= −2.
Then, using point–gradient form for a line with gradient −2 through A(1, 5), the line AB is y − y1 = m(x − x1 ) y − 5 = −2(x − 1) y − 5 = −2x + 2 y = −2x + 7.
Note: Using the coordinates of B(4, −1) rather than A(1, 5) would give the same equation.

WORKED EXERCISE:
Given the points A(6, 0) and B(0, 9), find, in general form, the equation of:
(a) the line AB,
(b) the line through A perpendicular to AB.

SOLUTION:
(a) First,

9−0
0−6
= −3 ,
2

gradient of AB =

so the line AB is

y = mx + b y= ×2

−3 x
2

+9

(This is gradient–intercept form.)
(The line has y-intercept 9.)

2y = −3x + 18
3x + 2y − 18 = 0.

(b) The gradient of the line perpendicular to AB is 2 ,
3
so the line is

y − y1 = m(x − x1 )T

(This is point–gradient form.)

y − 0 = 2 (x − 6)
3
×3

3y = 2(x − 6)
3y = 2x − 12
2x − 3y − 12 = 0.

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Intersection of Lines — Concurrent Lines: The point where two distinct lines intersect can be found using simultaneous equations, as discussed in Chapter One.
Three distinct lines are called concurrent if they all pass through the same point.

TESTING FOR CONCURRENT LINES: To test whether three given lines are concurrent:
• First find the point of intersection of two of them.

16

• Then test, by substitution, whether this point lies on the third line.

WORKED EXERCISE:
Test whether the following three lines are concurrent.
1

: 5x − y − 10 = 0,

2

: x + y − 8 = 0,

3

: 2x − 3y + 9 = 0.

SOLUTION:
A. Solve 1 and 2 simultaneously.
Adding 1 and 2 ,
6x − 18 = 0 x = 3, and substituting into 2 , 3 + y − 8 = 0 y=5 so the lines 1 and 2 intersect at (3, 5).
B. Substituting the point (3, 5) into the third line 3 ,
LHS = 6 − 15 + 9
=0
= RHS, so the three lines are concurrent, meeting at (3, 5).

Exercise 6D
1. Use point–gradient form y − y1 = m(x − x1 ) to find the equation of the line through the point P (3, 5) with gradient 3. Rearrange your answer into general form.
2. Use point–gradient form y − y1 = m(x − x1 ) to find the equation of the line through the point P (−2, 7) with each given gradient. Rearrange your answer into general form.
(a) gradient 6
(b) gradient −2
(c) gradient 2
(d) gradient − 7
3
2
3. Use point–gradient form y − y1 = m(x − x1 ) to find the equation of the line through each given point with gradient − 3 . Rearrange your answer into general form.
5
(a) through (1, 2)
(b) through (6, 0)
(c) through (−5, 3) (d) through (0, −4)
4. Find, in general form, the equation of the line:
(a) through (1, 1) with gradient 2,
(b) with gradient −1 through (3, 1),
(c) with gradient 3 through (−5, −7),
(d) through (0, 0) with gradient −5,
(e) through (−1, 3) with gradient − 1 ,
3
(f) with gradient − 4 through (3, −4).
5

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CHAPTER 6: Coordinate Geometry

6D Further Equations of Lines

159

5. Find, in gradient–intercept form, the equation of:
(a) the line through (2, 5) and parallel to y = 2x + 5,
(b) the line through (2, 5) and perpendicular to y = 2x + 5,
(c) the line through (5, −7) and parallel to y = −5x,
(d) the line through (5, −7) and perpendicular to y = −5x,
(e) the line through (−7, 6) and parallel to y = 3 x − 8,
7
(f) the line through (−4, 0) and perpendicular to y = − 2 x.
5
6. (a) Find the gradient of the line through the points A(4, 7) and B(6, 13).
(b) Hence use point–gradient form to find, in general form, the equation of the line AB.
7. Find the gradient of the line through each pair of points, and hence find its equation.
(g) (0, −1), (−4, 0)
(c) (−4, −1), (6, −6) (e) (−1, 0), (0, 2)
(a) (3, 4), (5, 8)
(h) (0, −3), (3, 0)
(f) (2, 0), (0, 3)
(b) (−1, 3), (1, −1) (d) (5, 6), (−1, 4)
8. (a) Find the gradient of the line through A(1, −2) and B(−3, 4).
(b) Hence find, in general form, the equation of:
(ii) the line through A and perpendicular to AB.
(i) the line AB,
9. Find the equation of the line parallel to 2x − 3y + 1 = 0 and:
(a) passing through (2, 2),
(b) passing through (3, −1).
10. Find the equation of the line perpendicular to 3x + 4y − 3 = 0 and:
(a) passing through (−1, −4),
(b) passing through (−2, 1).
DEVELOPMENT

11. (a) Find the point M of intersection of the lines 1 : x + y = 2 and 2 : 4x − y = 13.
(b) Show that M lies on 3 : 2x − 5y = 11, and hence that 1 , 2 and 3 are concurrent.
(c) Use the same method to test whether each set of lines is concurrent.
(i) 2x + y = −1, x − 2y = −18 and x + 3y = 15
(ii) 6x − y = 26, 5x − 4y = 9 and x + y = 9
12. Put the equation of each line in gradient–intercept form and hence write down the gradient.
Then find, in gradient–intercept form, the equation of the line that is:
(i) parallel to it through A(3, −1), (ii) perpendicular to it through B(−2, 5).
(c) 4x + 3y − 5 = 0
(a) 2x + y + 3 = 0
(b) 5x − 2y − 1 = 0
13. The angle of inclination α and a point A on a line are given below. Use the formula gradient = tan α to find the gradient of each line, then find its equation in general form.
(a) α = 45◦ , A = (1, 0)
(c) α = 30◦ , A = (4, −3)
(b) α = 120◦ , A = (−1, 0)
(d) α = 150◦ , A = (−2, −5)
14. Determine, in general form, the equation of each straight line sketched below.
(a)
(b)
(c)
(d) y y

(i) x (ii)

(3,−1)

y

y

(i)

3

x

−6

4
(ii)
2

x
45º

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CHAPTER 6: Coordinate Geometry

CAMBRIDGE MATHEMATICS 2 UNIT YEAR 11

15. Explain why the four lines 1 : y = x + 1, 2 : y = x − 3, 3 : y = 3x + 5 and enclose a parallelogram. Then find the vertices of this parallelogram.

4:

r

y = 3x − 5

16. Show that the triangle with vertices A(1, 0), B(6, 5) and C(0, 2) is right-angled. Then find the equation of each side.
17. The three points A(1, 0), B(0, 8) and C(7, 4) form a triangle.
Let θ be the angle between AC and the x-axis.
(a) Find the gradient of the line AC and hence determine θ, correct to the nearest degree.
(b) Derive the equation of AC.
(c) Find the coordinates of the midpoint D of AC.
(d) Show that AC is perpendicular to BD.
(e) What type of triangle is ABC?
(f) Find the area of this triangle.
(g) Write down the coordinates of a point E such that the parallelogram ABCE is a rhombus.
18. (a)
(b)
(c)
(d)

y
B(0,8)
C(7,4)
D

θ x A(1,0)

On a number plane, plot the points A(4, 3), B(0, −3) and C(4, 0).
Find the equation of BC.
Explain why OABC is a parallelogram.
Find the area of OABC and the length of the diagonal AB.

19. The line crosses the x- and y-axes at L(−4, 0) and M (0, 3).
The point N lies on and P is the point P (0, 8).
(a) Copy the sketch and find the equation of .
(b) Find the lengths of ML and MP and hence show that
LMP is an isosceles triangle.
(c) If M is the midpoint of LN , find the coordinates of N .
(d) Show that NPL = 90◦ .
(e) Write down the equation of the circle through N , P and L.
20. The vertices of a triangle are P (−1, 0) and Q(1, 4) and R, where R lies on the x-axis and QP R = QRP = θ.
(a) Find the coordinates of the midpoint of P Q.
(b) Find the gradient of P Q and show that tan θ = 2.
(c) Show that P Q has equation y = 2x + 2.

y
P(0,8)

l
N
M(0,3)

L(−4,0)

x

y
Q(1,4)

P(−1,0)

(d) Explain why QR has gradient −2, and hence find its equation. θ (e) Find the coordinates of R and hence the area of triangle P QR.
(f) Find the length QR, and hence find the perpendicular distance from P to QR.

R x CHALLENGE

21. Find k if the lines 1 : x + 3y + 13 = 0, 2 : 4x + y − 3 = 0 and 3 : kx − y − 10 = 0 are concurrent. [Hint: Find the point of intersection of 1 and 2 and substitute into 3 .]
22. Consider the two lines 1 : 3x + 2y + 4 = 0 and 2 : 6x + μy + λ = 0.
(a) Write down the value of μ if: (i) 1 is parallel to 2 , (ii) 1 is perpendicular to
(b) Given that 1 and 2 intersect at a point, what condition must be placed on μ?
(c) Given that 1 is parallel to 2 , write down the value of λ if:
(i) 1 is the same line as 2 ,
(ii) the distance between the y-intercepts of the two lines is 2.

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CHAPTER 6: Coordinate Geometry

6E Perpendicular Distance

161

6 E Perpendicular Distance
The shortest distance from a point P to a line is the perpendicular distance, which is the distance p in the diagram to the right. There is a simple formula for this perpendicular distance, which is proven in the appendix to this chapter.

17

PERPENDICULAR DISTANCE FORMULA:
The perpendicular distance p from the point (x1 , y1 ) to the line ax + by + c = 0 is p= |ax1 + by1 + c|
√
. a2 + b2

y

P(x1,y1) p l

x

O

Notice that the numerator is the point (x1 , y1 ) substituted into |ax + by + c|. The line must be rearranged into general form before the formula can be applied.

WORKED EXERCISE:

Find the perpendicular distance from the point P (−2, 5) to the line y = 2x − 1.

SOLUTION:
The line in general form is 2x − y − 1 = 0,
|ax1 + by1 + c|
√
so perpendicular distance = a2 + b2
| − 4 − 5 − 1|
=
22 + (−1)2
√
5
10
=√ ×√
5
5
√
10 5
=
5
√
= 2 5.

(Substitute P (−2, 5) into 2x − y − 1.)
(Rationalise the denominator.)

Distance between Parallel Lines: To find the distance between two parallel lines, choose any point on one line and find its perpendicular distance from the other line.

WORKED EXERCISE:
Find the perpendicular distance between the two parallel lines
2x + 5y − 1 = 0

and

2x + 5y − 7 = 0.

SOLUTION:
Choose any convenient point on the first line, say P (3, −1). (Its coefficients are integers.)
The distance between the lines is the perpendicular distance from P to the second line, y |ax1 + by1 + c|
√
so perpendicular distance =
2x + 5y − 7 = 0 a2 + b2
|6 − 5 − 7|
= √
22 + 52
| − 6|
=√
x
4 + 25
2x + 5y − 1 = 0
6
P(3,−1)
=√ .
29

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CHAPTER 6: Coordinate Geometry

CAMBRIDGE MATHEMATICS 2 UNIT YEAR 11

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Circles and the Perpendicular Distance Formula: A line is a tangent to a circle when its perpendicular distance from the centre is equal to the radius. Lines closer to the centre are secants, and lines more distant miss the circle entirely.

WORKED EXERCISE:
(a) Show that : 3x+4y −20 = 0 is a tangent to the circle (x−7)2 +(y −6)2 = 25.
(b) Find the length of the chord of the circle cut off by the line m: 3x+4y−60 = 0.

SOLUTION:
The circle has centre (7, 6) and radius 5.
(a) The distance from the line to the centre is
|21 + 24 − 20| p= √
32 + 42
= 5, so is a tangent to the circle.
(b) The distance p from the line m to the centre is
|21 + 24 − 60| p= √
32 + 42
= 3.
Using Pythagoras’ theorem in the circle to the right, chord length = 2 × 4
= 8 units.

y

3x + 4y = 60
4
4

3
5

5

3x + 4y = 20

x

WORKED EXERCISE:

For what values of k will the line 5x − 12y + k = 0 never intersect the circle with centre P (−3, 1) and radius 6?

SOLUTION:
The condition is

|ax1 + by1 + c|
√
> 6 (The perpendicular distance exceeds the radius.) a2 + b2
| − 15 − 12 + k|
√
>6
52 + 122
| − 27 + k|
>6
13
|k − 27| > 78 k < −51 or k > 105.

Exercise 6E
|ax1 + by1 + c|
√
to find the perpendicular distance from each line to the origin. a2 + b2
(a) x + 3y + 5 = 0
(b) 2x − y + 4 = 0
(c) 2x + 4y − 5 = 0

1. Use p =

|ax1 + by1 + c|
√
to find the perpendicular distance from each point to each line. a2 + b2
(d) (−3, −2) and x + 3y + 4 = 0
(a) (2, 0) and 3x + 4y − 1 = 0
(b) (−2, 1) and 12x − 5y + 3 = 0
(e) (3, −1) and x + 2y − 1 = 0
(c) (−3, 2) and 4x − y − 3 = 0
(f) (1, 3) and 2x + 4y + 1 = 0

2. Use p =

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6E Perpendicular Distance

163

3. Which of the given points is: (a) closest to, (b) furthest from, the line 6x−8y −9 = 0?
A(1, −1)
B(3, 2)
C(−4, 1)
D(−3, −3)
4. Which of the given lines is: (a) closest to, (b) furthest from, the point (−1, 5)?
1 : 2x + 3y + 4 = 0
2 : x − 4y + 7 = 0
3 : 3x + y − 8 = 0
DEVELOPMENT

5. Write down the centre and radius of each circle. Then use the perpendicular distance formula to determine how many times the given line intersects the circle.
(a) 3x − 5y + 16 = 0, x2 + y 2 = 5
(c) 3x − y − 8 = 0, (x − 1)2 + (y − 5)2 = 10
2
2
(b) 7x + y − 10 = 0, x + y = 2
(d) x + 2y + 3 = 0, (x + 2)2 + (y − 1)2 = 6
6. Choose any point on the first line to find the distance between the parallel lines in each pair. (b) 4x + y − 2 = 0, 4x + y + 8 = 0
(a) x − 3y + 5 = 0, x − 3y − 2 = 0
√
7. (a) The line y − 2x + k = 0 is 2 5 units from the point (1, −3). Find k.
(b) The line 3x − 4y + 2 = 0 is 3 units from the point (−1, ). Find .
5
8. Find the range of values that k may take if:
1
(a) the line y − x + k = 0 is more than √2 units from the point (2, 7).
√
(b) the line x + 2y − 5 = 0 is at most 5 units from the point (k, 3).
9. The vertices of a triangle are A(−3, −2), B(3, 1) and C(−1, 4).
(a) Find the equation of the side AB in general form.
(b) How far is C from this line?
(c) Find the length of AB and hence find the area of this triangle.
(d) Similarly, find the area of the triangle with vertices P (1, −1), Q(−1, 5) and R(−3, 1).
10. Draw on a number plane the triangle ABC with vertices A(5, 0), B(8, 4) and C(0, 10).
(a) Show that the line AB has equation 3y = 4x − 20.
(b) Show that the gradient of BC is − 3 .
4
(c) Hence show that AB and BC are perpendicular.
(d) Show that the interval AB has length 5 units.
(e) Show that the triangles AOC and ABC are congruent.
(f) Find the area of quadrilateral OABC.
(g) Find the distance from D(0, 8) to the line AB.
CHALLENGE

y

11. (a) Write down the centre and radius of the circle with equation (x + 2)2 + (y + 3)2 = 4. Then find the distance from the line 2y − x + 8 = 0 to the centre.
(b) Use Pythagoras’ theorem to determine the length of the chord cut off from the line by the circle.

x

12. (a) Write down the equation of a line through the origin with gradient m.
(b) Write down the distance from this line to the point (3, 1).
(c) If the line is tangent to the circle (x − 3)2 + (y − 1)2 = 4, show that m satisfies the equation 5m2 − 6m − 3 = 0.
(d) Find the possible values of m, and hence find the equations of the two tangents.

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6 F Lines Through the Intersection of Two Given Lines
This section develops an ingenious way of finding the equations of particular lines through the intersection of two given lines without actually solving the two given lines simultaneously to find their point of intersection. The method is another situation where the general form of the equation of the line is used.

The General Form of Such a Line: Suppose that two lines
1 : a1 x + b1 y + c1 = 0

and

y

l

l2

2 : a2 x + b 2 y + c2 = 0

l1

intersect at a point M . The set of all the lines through M forms a family of lines through M . In the diagram to the right, the line is one of these lines.

M

It is proven in the appendix to this chapter that every line passing through M has the form
(a1 x + b1 y + c1 ) + k(a2 x + b2 y + c2 ) = 0

x

(∗)

for some value of the constant k. Notice that substituting different values of k into (∗) will clearly yield different lines — the proof in the appendix simply shows that all these lines pass through M .

LINE THROUGH THE INTERSECTION OF TWO GIVEN LINES:
Let two lines 1 : a1 x + b1 y + c1 = 0 and 2 : a2 x + b2 y + c2 ) = 0 intersect at a
18 point M . Then every line passing through M has the form
(a1 x + b1 y + c1 ) + k(a2 x + b2 y + c2 ) = 0, where k is a constant.
To find any particular line through M , all that is necessary is to substitute the extra information into the equation above to find the value of k

WORKED EXERCISE:
Find the equation of the line that passes through the intersection M of the lines
1:

x + 2y − 6 = 0

and

2:

3x − 2y − 6 = 0

and also passes through the point P (2, −1).

SOLUTION:
Using Box 18, any line through M has equation
(x + 2y − 6) + k(3x − 2y − 6) = 0, for some constant k.

(1)

Substituting the point P (2, −1) into equation (1) gives
(2 − 2 − 6) + k(6 + 2 − 6) = 0
−6 + 2k = 0
2k = 6 k = 3.
Substituting k = 3 back into equation (1), the required line is
(x + 2y − 6) + 3(3x − 2y − 6) = 0 x + 2y − 6 + 9x − 6y − 18 = 0
10x − 4y − 24 = 0
5x − 2y − 12 = 0.

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6F Lines Through the Intersection of Two Given Lines

165

WORKED EXERCISE:
Find the equation of the line that passes through the intersection M of the lines x + 2y − 6 = 0

1:

(a) and has gradient 5,

and

2:

3x − 2y − 6 = 0

(b) and is vertical.

SOLUTION:
As in the previous worked exercise, any line through M has equation
(x + 2y − 6) + k(3x − 2y − 6) = 0, for some constant k.

(1)

Equation (1) can be rearranged into general form by collecting terms in x, and in y.
(x + 2y − 6) + k(3x − 2y − 6) = 0 x + 2y − 6 + 3kx − 2ky − 6k = 0
(1 + 3k)x + (2 − 2k)y − (6 + 6k) = 0
(2)
(a) Now making y the subject, (2 − 2k)y = (−1 − 3k)x + (6 + 6k)
(−1 − 3k)x
6 + 6k y= +
2 − 2k
2 − 2k
−1 − 3k and so the gradient of the line with equation (1) is
.
2 − 2k
−1 − 3k
=5
Since the gradient is 5,
2 − 2k
−1 − 3k = 10 − 10k
7k = 11 k = 11 .
7
Substituting back into equation (1), the required line is
(x + 2y − 6) + 11 (3x − 2y − 6) = 0
7
7(x + 2y − 6) + 11(3x − 2y − 6) = 0
40x − 8y − 108 = 0
10x − 2y − 27 = 0.
(b) Since the line is vertical, the coefficient of y in equation (2) is zero. that is, 2 − 2k = 0 k = 1.
Hence from (2), the required line is 4x − 12 = 0 x = 3.

Exercise 6F
1. (a) Graph the lines x − y = 0 and x + y − 2 = 0 on grid or graph paper and label them
(they intersect at (1, 1)).
(b) Simplify the equation (x − y) + k(x + y − 2) = 0 for k = 2, 1, 1 and 0. Add these lines
2
to your graph, and label each line with its value of k. Observe that each line passes through (1, 1).
(c) Repeat this process for k = − 1 , −1 and −2, adding these lines to your graph.
2
2. The lines x + 2y + 9 = 0 and 2x − y + 3 = 0 intersect at B.
(a) Write down the general equation of a line through B.
(b) Hence find the equation of the line through B and the origin O.

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r

3. (a) Write down the general form of a line through the point T of intersection of the two lines 2x − 3y + 6 = 0 and x + 3y − 15 = 0.
(b) Hence find the equation of the line through T and:
(iii) (−3, 3)
(iv) (0, 0)
(i) (3, 8)
(ii) (6, 0)
4. Find the equation of the line through the intersection of the lines x − y − 3 = 0 and y + 3x − 5 = 0 and each given point. Do not find the point of intersection of the lines.
(a) (0, −2)
(b) (−1, 5)
(c) (3, 0)
5. The lines 2x + y − 5 = 0 and x − y + 2 = 0 intersect at A.
(a) Write down the general equation of a line through A, and show that it can be written in the form x(2 + k) + y(1 − k) + (2k − 5) = 0.
(b) Find the value of k that makes the coefficient of x zero, and hence find the equation of the horizontal line through A.
(c) Find the value of k that makes the coefficient of y zero, and hence find the equation of the vertical line through A.
(d) Hence write down the coordinates of A.
DEVELOPMENT

6. (a) The general form of a line through the intersection M of x−2y+5 = 0 and x+y+2 = 0
1+k
is : (x − 2y + 5) + k(x + y + 2) = 0. Show that the gradient of is
.
2−k
(b) Hence find the equation of the line that passes through M and is:
(iii) perpendicular to 5y − 2x = 4,
(i) parallel to 3x + 4y = 5,
(ii) perpendicular to 2x − 3y = 6,
(iv) parallel to x − y − 7 = 0.
7. (a) Show that the three lines 1 : 2x − 3y + 13 = 0, 2 : x + y − 1 = 0 and 3 : 4x + 3y − 1 = 0 are concurrent by the following method.
(i) Without finding any points of intersection, find the equation of the line that passes through the intersection of 1 and 2 and is parallel to 3 .
(ii) Show that this line is the same line as 3 .
(b) Use the same method as in the previous question to test each family of lines for concurrency. (i) 1 : 2x − y = 0, 2 : x + y = 9 and 3 : x − 3y + 15 = 0
(ii) 1 : x + 4y + 6 = 0, 2 : x + y − 3 = 0 and 3 : 7x − 3y − 10 = 0
8. (a) Find the point P of intersection of x + y − 2 = 0 and 2x − y − 1 = 0.
(b) Show that P satisfies the equation x + y − 2 + k(2x − y − 1) = 0.
(c) Find the equation of the line through P and Q(−2, 2):
(i) using the coordinates of both P and Q,
(ii) without using the coordinates of P .
The two answers should be the same.
CHALLENGE

9. (a) It is known that the line : x + 2y + 10 = 0 is tangent to the circle C: x2 + y 2 = 20 at the point T . Use the fact that a tangent is perpendicular to the radius at the point of contact to write down the equation of the radius OT of the circle.
(b) Without actually finding the coordinates of T , use the result of part (a) to find the equation of the line through S(1, −3) and the point of contact T .

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167

6 G Coordinate Methods in Geometry
When the French mathematician and philosopher Ren´ Descartes introduced coe ordinate geometry in the 17th century, he intended it as a method of proving all the theorems of Euclidean geometry using algebraic rather than geometric arguments. The questions in this exercise apply coordinate methods to proving some well-known geometric theorems.
Some questions use the words altitude and median.

19

MEDIAN: A median of a triangle is the interval from a vertex of the triangle to the midpoint of the opposite side.
ALTITUDE: An altitude of a triangle is the perpendicular from a vertex of the triangle to the opposite side (extended if necessary).

Example — The Three Altitudes of a Triangle are Concurrent: There are three altitudes in a triangle — the theorem below asserts the concurrence of these three altitudes.
The proof begins by placing the triangle carefully on the coordinate plane so that two points lie on the x-axis and the third lies on the y-axis. Any triangle can be placed in this way by rotating and shifting it appropriately, so the final theorem applies to any triangle.
Theorem: The three altitudes of a triangle are concurrent. (Their point of intersection is called the orthocentre of the triangle.)
Proof: Let the side AB of the triangle ABC lie on the x-axis, and the vertex
C lie on the positive side of the y-axis. Let the coordinates of the vertices be
A(a, 0), B(b, 0) and C(0, c), as in the diagram.
First, the altitude through C is the interval CO on the y-axis.

y

Secondly, let M be the foot of the altitude through A. c b
Since BC has gradient − , AM has gradient (see Box 9), b c b so the equation of AM is y − 0 = (x − a) (See Box 14.) c bx ab
−
y= c c and substituting x = 0, the altitude AM meets CO at V

0, −

C(0,c)
M

V
A(a,0) O

B(b,0) x

ab
.
c

Thirdly, let N be the foot of the altitude through B (not shown on the diagram). c a
Since AC has gradient − , BN has gradient (see Box 9), a c a so the equation of BN is y − 0 = (x − b) (See Box 14.) c ax ab
−
y= c c ab and substituting x = 0, the altitude BN meets CO at the same point V 0, −
.
c
Hence the three altitudes are concurrent.

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Exercise 6G
Note:

Diagrams should be drawn wherever possible.

1. (a) On a number plane, plot the points O(0, 0), A(6, 0), B(6, 6) and C(0, 6), which form a square.
(i) Find the gradients of the diagonals OB and AC.
(ii) Hence show that the diagonals OB and AC are perpendicular.
(b) Theorem: The diagonals of a square are perpendicular.
Let the vertices of the square be O(0, 0), A(a, 0), B(a, a) and C(0, a).
(i) Find the gradients of the diagonals OB and AC.
(ii) Hence show that the diagonals OB and AC are perpendicular.
2. (a) The points O(0, 0), P (8, 0) and Q(0, 10) form a rightangled triangle. Let M be the midpoint of P Q.
(i) Find the coordinates of M .
(ii) Find the distances OM , P M and QM , and show that M is equidistant from each of the vertices.
(iii) Explain why a circle with centre M can be drawn through the three vertices O, P and Q

y
Q(0,2q)
M x P(2p,0)

O

(b) Theorem: The midpoint of the hypotenuse of a right-angled triangle is the centre of a circle through all three vertices.
Prove this theorem for any right-angled triangle by placing its vertices at O(0, 0),
P (2p, 0) and Q(0, 2q) and repeating the procedures of part (a).
3. (a) Let P QRS be the quadrilateral with vertices P (1, 0), Q(0, 2), R(−3, 0) and S(0, −4).
(i) Find P Q2 , RS 2 , P S 2 and QR2 .
(ii) Show that P Q2 + RS 2 = P S 2 + QR2 .
(b) Theorem: If the diagonals of a quadrilateral are perpendicular, then the two sums of squares of opposite sides are equal.
Prove this theorem for any quadrilateral by placing the vertices on the axes, giving them coordinates P (p, 0), Q(0, q), R(−r, 0) and S(0, −s), and proceeding as in part (a).
4. (a) A triangle has vertices at A(1, −3), B(3, 3) and C(−3, 1).
(i) Find the coordinates of the midpoint P of AB and the midpoint Q of BC.
(ii) Show that P Q AC and that P Q = 1 AC.
2
(b) Theorem: The interval joining the midpoints of two sides of a triangle is parallel to the base and half its length.
Prove this theorem for any triangle by placing its vertices at A(2a, 0), B(2b, 2c) and
C(0, 0), where a > 0, and proceeding as in part (a).
DEVELOPMENT

5. Theorem: The midpoints of the sides of a quadrilateral form a parallelogram.
Let the vertices of the quadrilateral be A(a1 , a2 ), B(b1 , b2 ),
C(c1 , c2 ) and D(d1 , d2 ), as in the diagram opposite.
(a) Find the midpoints P , Q, R and S of the sides AB, BC,
CD and DA respectively. (The figure P QRS is also a quadrilateral.) (b) Find the midpoints of the diagonals P R and QS.
(c) Explain why this proves that P QRS is a parallelogram.

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169

6. The points A(a, 0) and Q(q, 0) are points on the positive x-axis, and the points B(0, b) and P (0, p) lie on the positive y-axis. Show that AB 2 − AP 2 = QB 2 − QP 2 .
7. The triangle OBA has its vertices at the origin O(0, 0), A(3, 0) and B(0, 4). The point C lies on AB, and OC is perpendicular to AB. Draw a diagram showing this information.
(a) Find the equations of AB and OC and hence find the coordinates of C.
(b) Find the lengths OA, AB, OC, BC and AC.
(c) Thus confirm these important results for a right-angled triangle:
(i) OC 2 = AC × BC
(ii) OA2 = AC × AB
8. The diagram opposite shows the points A, B, C and D on the number plane.
(a) Show that ABC is equilateral.
(b) Show that ABD is isosceles, with AB = AD.
(c) Show that AB 2 = 1 BD2 .
3

y
B a 3

C
−a

A a Dx
3a

9. Theorem: The diagonals of a parallelogram bisect each other.
Place three vertices of the parallelogram at A(0, 0), B(2a, 2b) and D(2c, 2d).
(a) Use gradients to show that with C = (2a + 2c, 2b + 2d), the quadrilateral ABCD is a parallelogram. (b) Find the midpoints of the diagonals AC and BD.
(c) Explain why this proves that the diagonals bisect each other.
CHALLENGE

10. (a) The points A(1, −2), B(5, 6) and C(−3, 2) are the vertices of a triangle, and P , Q and R are the midpoints of BC, CA and AB respectively.
(i) Find the equations of the three medians BQ, CR and AP .
(ii) Find the intersection of BQ and CR, and show that it lies on the third median AP .
(b) Theorem: The three medians of a triangle are concurrent. (Their point of intersection is called the centroid.)
Prove that the theorem is true for any triangle by choosing as vertices A(6a, 6b),
B(−6a, −6b) and C(0, 6c), and following these steps.
(i) Find the midpoints P , Q and R of BC, CA and AB respectively.
(ii) Show that the median through C is x = 0 and find the equations of the other two medians. (iii) Find the point where the median through C meets the median through A, and show that this point lies on the median through B.
11. Theorem: The perpendicular bisectors of the sides of a triangle are concurrent. Their point of intersection (called the circumcentre) is the centre of a circle through all three vertices (called the circumcircle).
Prove this theorem for any triangle by placing its vertices at A(2a, 0), B(−2a, 0) and
C(2b, 2c), and proceeding as follows.
(a) Find the gradients of AB, BC and CA.
(b) Hence find the equations of the three perpendicular bisectors.
(c) Find the intersection M of any two, and show that it lies on the third.
(d) Explain why the circumcentre must be equidistant from each vertex.

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6H Chapter Review Exercise
1. Let X = (2, 9) and Y = (14, 4). Use the standard formulae to find:
(a) the midpoint of XY ,
(b) the gradient of XY ,
(c) the length of XY .
2. A triangle has vertices A(1, 4), B(−3, 1) and C(−2, 0).
(a) Find the lengths of all three sides of ABC.
(b) What sort of triangle is ABC?
3. A quadrilateral has vertices A(2, 5), B(4, 9), C(8, 1) and D(−2, −7).
(a) Find the midpoints P of AB, Q of BC, R of CD and S of DA.
(b) Find the gradients of P Q, QR, RS and SP .
(c) What sort of quadrilateral is P QRS?
4. A circle has diameter AB, where A = (2, −5) and B = (−4, 7).
(a) Find the centre C and radius r of the circle.
(b) Use the distance formula to test whether P (6, −1) lies on the circle.
5. (a) Find the gradients of the sides of LM N , given L(3, 9), M (8, −1) and N (−1, 7).
(b) Explain why LM N is a right-angled triangle.
6. (a)
(b)
(c)
(d)

Find
Find
Find
Find

the gradient of the interval AB, where A = (3, 0) and B = (5, −2). a if AP ⊥ AB, where P = (a, 5). the point Q(b, c) if B is the midpoint of AQ. d if the interval AD has length 5, where D = (6, d).

7. Find, in general form, the equation of the line:
(a) with gradient −2 and y-intercept 5,
(b) with gradient 2 through the point A(3, 5),
3
(c) through the origin perpendicular to y = 7x − 5,
(d) through B(−5, 7) parallel to y = 4 − 3x,
(e) with y-intercept −2 and angle of inclination 60◦ .
8. Put the equation of each line in gradient–intercept form, and hence find its y-intercept b, its gradient m and its angle of inclination α (correct to the nearest minute when necessary).
(a) 5x − 6y − 7 = 0
(b) 4x + 4y − 3 = 0
9. Find the gradient of each line AB, then find its equation in general form.
(b) A(5, −2) and B(7, −7)
(a) A(3, 0) and B(4, 8)
10. (a) Are the points L(7, 4), M (13, 2) and N (25, −3) collinear?
(b) Are the lines 2x + 5y − 29 = 0, 4x − 2y + 2 = 0 and 7x − 3y + 1 = 0 concurrent?
11. (a) Determine whether the lines 8x + 7y + 6 = 0, 6x − 4y + 3 = 0 and 2x + 3y + 9 = 0 enclose a right-angled triangle.
(b) Determine what sort of figure the lines 4x + 8y + 3 = 0, 5x − 2y + 7 = 0, x + 2y − 6 = 0 and 9x − 3y = 0 enclose.
12. (a) Find the points where the line 5x + 4y − 30 = 0 meets the x-axis and y-axis.
(b) Hence find the area of the triangle formed by the line, the x-axis and the y-axis.

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13. Consider
(a) Find
(b) Find
(c) Find

the line : kx + 3y + 12 = 0, k if passes through (2, 4). k if has x-intercept −36. k if has gradient 9.

6H Chapter Review Exercise

171

where k is a constant.
(d) Find k if is parallel to 6x − y − 7 = 0.
(e) Find k if is perpendicular to 6x − y − 7 = 0.
(f) Explain why can never have y-intercept 1.

14. A sketch is essential in this question.
(a) Find the gradient, length and midpoint M of the interval joining A(10, 2) and B(2, 8).
(b) Show that the perpendicular bisector of the interval AB has equation 4x − 3y − 9 = 0.
(c) Find the point C where the perpendicular bisector meets the line x − y + 2 = 0.
(d) Use the distance formula to show that C is equidistant from A and B.
(e) Show that CM = 15 and hence find the area of ABC.
(f) Let θ = ACB. Use the area formula area = 1 × AC × BC × sin θ to find θ, correct
2
to the nearest minute.
15. (a) Find the perpendicular distance from A(−2, −3) to the line 5x + y + 2 = 0.
(b) Find the distance between the parallel lines 3x + y − 3 = 0 and 3x + y + 7 = 0.
[Hint: Choose a point on one line and find its perpendicular distance from the other.]
(c) Find k if : 3x + 4y + 3 = 0 is a tangent to the circle with centre A(k, −5) and radius 3.
[Hint: This means that the perpendicular distance from A to is 3.]
16. A triangle has vertices R(3, −4), S(−6, 1) and T (−2, −2).
(a) Find the gradient of the line ST , and hence find its equation.
(b) Find the perpendicular distance from R to ST .
(c) Find the length of the interval ST and hence find the area of

RST .

17. Draw a sketch of the quadrilateral with vertices J(2, −5), K(−4, 3), L(4, 9) and M (13, −3).
(a) Show that JKLM is a trapezium with JK M L.
(b) Find the lengths of the parallel sides JK and M L.
(c) Find the equation of the line JK.
(d) Find the perpendicular distance from M to JK.
(e) Hence find the area of the trapezium JKLM .
18. Let M be the point of intersection of the lines 1 : 3x − 4y − 5 = 0 and 2 : 4x + y + 7 = 0.
Write down the equation of the general line through M . Hence, without actually finding the coordinates of M , find the equation of:
(a) the line through M and A(1, 1),
(c) the vertical line through M ,
(b) the line through M with gradient −3,
(d) the horizontal line through M .
19. Theorem: The angle subtended by a diameter of a circle at any point on the circumference of the circle is a right angle.
Let the circle have centre the origin O and radius r. y r
Let the diameter AB lie on the x-axis.
P(a,b)
Let A = (r, 0) and B = (−r, 0).
Let P (a, b) be any point on the circle.
B
A
-r
r x
(a) Find P O2 and hence explain why a2 + b2 = r2 .
(b) Find the gradients of AP and BP .
-r
b2
(c) Show that the product of these gradients is 2
.
a − r2
(d) Use parts (a) and (c) to show that AP ⊥ BP .

ISBN: 9781107679573
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172

CHAPTER 6: Coordinate Geometry

CAMBRIDGE MATHEMATICS 2 UNIT YEAR 11

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Appendix — The Proofs of Two Results
This appendix contains proofs of two results used in the chapter — the perpendicular distance formula and the equation of the general line through the intersection of two given lines.

The Perpendicular Distance Formula: The formula was stated in Section 6E.
Theorem: Let p be the perpendicular distance from the point P (x1 , y1 ) to the line with equation ax + by + c = 0.

y

P(x1,y1)

|ax1 + by1 + c|
√
Then p =
.
a2 + b2

Perpendicular Distance from the Origin: The first step in the proof is to develop a formula for the perpendicular distance p of a given line : ax + by + c = 0 from the origin.
Consider the triangle OAB formed by the line

l

p

x

O

and the two axes.

We find two different expressions for the area of

OAB and equate them.

First, substituting y = 0, and then x = 0, shows that the line ax + by + c = 0 c c has x-intercept − and y-intercept − , a b and so, using OA as the base of OAB,
OAB =

1
2

=

1
2

=

area of

1
2

× OA × OB c c
× − × − a b
2
c
.
ab

Secondly, the side AB is the hypotenuse of

y
B

OAB,

c2 c2 + 2 a2 b
2 2 c (a + b2 )
=
, a2 b2 c AB = a2 + b2 , ab and so, using AB as the base of OAB, c area of OAB = 1 p a2 + b2 .
2
ab so by Pythagoras’ theorem, AB 2 =

Equating the two expressions for the area of
1
2p

c ab p

l p O

A

x

ABO,

2

a2 + b2 =

c ab 1
2

a2 + b2 = |c| p= √

|c|
.
+ b2

a2

ISBN: 9781107679573
© Bill Pender, David Sadler, Julia Shea, Derek Ward 2012
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CHAPTER 6: Coordinate Geometry

Appendix — The Proofs of Two Results

y

Completion of Proof: Now we can use shifting to find the perpen-

P(x1,y1) p dicular distance p from any point P (x1 , y1 ) to the line

173

l

: ax + by + c = 0. l' The perpendicular distance remains the same if we shift both line and point x1 units to the left and y1 units down.

x

p

O

This shift moves the point P to the origin, and moves the line to the new line with equation a(x + x1 ) + b(y + y1 ) + c = 0

(Replace x by x + x1 and y by y + y1 .)

Rearranging the equation into general form, ax + by + (ax1 + by1 + c) = 0.
Then, using the formula previously established for the distance from the origin, the perpendicular distance p from the point (x1 , y1 ) to the line ax + by + c = 0 is p= |ax1 + by1 + c|
√
. a2 + b2

Lines Through the Intersection of Two Given Lines:
Section 6F gave the general form of the equation of any line through the intersection of two given lines. Here is the proof of that formula.
Theorem: Let two lines 1 : a1 x + b1 y + c1 = 0 and 2 : a2 x + b2 y + c2 ) = 0 intersect at a point M . Then every line passing through M has the form
(a1 x + b1 y + c1 ) + k(a2 x + b2 y + c2 ) = 0

(∗)

bp where k is a constant.
Proof: For all values of k, equation (∗) above is the equation of a line, and different values of k give different lines. y The proof below simply shows that the line with equation (∗) always passes through M , whatever the value of k.

l2

Let the intersection of 1 and 2 have coordinates M (x0 , y0 ).
We know that the first line 1 passes through M , so, substituting the coordinates of M into the equation of a1 x0 + b1 y0 + c1 = 0.

l l1 M(x0,y0)

1,

(1)

x

Similarly, we know that the second line 2 passes through M , so, substituting the coordinates of M into the equation of 2 , a2 x0 + b2 y0 + c2 = 0.
To prove that

(2)

passes through M , substitute M (x0 , y0 ) into (∗):

LHS = (a1 x0 + b1 y0 + c1 ) + k(a2 x0 + b2 y0 + c2 )
= 0 + 0, by the identities (1) and (2)
= RHS, as required.

Online Multiple Choice Quiz

ISBN: 9781107679573
© Bill Pender, David Sadler, Julia Shea, Derek Ward 2012
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