THE STRAIGHT LINE
A-GRADIENTS
The steepness of a line is described by its gradient. The gradient of the straight line joining
A(x1 , y1 ) and B(x2 , y2 ) is denoted by mAB where
Change in y
y
► The gradient of AB = mAB = Change in x
B(x2 , y2 )
y − y1
= 2
x2 − x1
y2 − y1
A(x1 , y1 )
x2 − x1
O
Example 1:
Find the gradient of the through A(3, -2) and B(6, 4).
SOLUTION
y −y
Gradient = x2 − x1
2
=
1
(x1 , y1 ) = A(3, − 2) and (x2 , y2 ) = B(6, 4)
4−(−2)
6
6−3
=
3
=2
Example 2:
value of k.
If the gradient of the line joining the points (k, 5) and (2, k) is −2, find the
SOLUTION
y −y
Gradient = x2 − x1
2
k−5
k−5
1
= 2−k
= −2
k − 5 = −2(2 − k)
k − 5 = −4 + 2k
−5 + 4 = 2k − k
k = −1
2−k
(x1 , y1 ) = (k, 5) and (x2 , y2 ) = B(2, k)
x
EXERCISE
1. Find the gradient of the line passing through each of the following pairs of points.
(a) A(2, 3) and B(7, 5)
(b) P( − 2, 8) and Q(1, -1)
(c) C(3, 1) and D(6, 3)
(d) M( − 4, 1) and N(16, 1)
(e) P( − 3k, − 4k) and Q( − k, 6k)
(f) A(5w, − 4w) and B(7w, 8w)
2. If the gradient of the line joining the points (4, − 9) and ( − 3, h) is −3, find the value of h.
3
3. If the gradient of the line joining the points ( − 3, − 7) and (4, y) is 5, find the value of y.
4. Find the gradient of line joining the points (2b, − 5b) and (6b, 3b).
5. The coordinates of A and B are (3k, 8) and (k, − 3) respectively. Given that the gradient of
the line segment AB is 3, find the value of k.
6. Find the gradient of the line that passes through the points P( − 3, 1) and Q(4, 5).
y
Q(4, 5)
P(−3, 1)
x
O
GRADIENT OF VERTICAL HORIZONTAL LINES
Example 6:
Find the gradients of the following straight lines:
y
P(−2, 3)
Q(2, 3)
A(4, 1)
x
O
B(4, − 2)
SOLUTIONS
3−3
Gradient of PQ = 2−(−2)
0
The line PQ is parallel to the x-axis.
=4
=0
The gradient of any line parallel to the x-axis is zero.
Gradient of AB =
3
1−(−2)
The line AB is parallel to the y-axis.
4−4
=0
The gradient of any line parallel to the y-axis is undefined.
B-GRADIENTS OF PARALLEL LINES
► Two lines l1 and l2 with gradients m1 and m2 are parallel to each other and their gradients
y
are equal.
m 1 = m2
l1
l2
x
O
Example 2:
Given the points A(3, 6), B(7, -2), C(4, -5) and D( − 1, 5), show that the straight
lines AB and CD are parallel to each other.
SOLUTION
y −y
Gradient of AB = x2 − x 1
2
(x1 , y1 ) = A(3, 6) and (x2 , y2 ) = B(7, − 2)
1
−2 − 6
= 7−3
−8
= 4
= −2
y −y
Gradient of CD = x2 − x 1
2
1
(x1 , y1 ) = C(4, − 5) and (x2 , y2 ) = D( − 1, 5)
5 − (−5)
= −1− 4
10
= −5
= −2
Since the gradients are equal, the straight lines AB and CD are parallel.
EXERCISE
1. Given the points P(3, 4), Q(6, 8), R( − 2, − 5) and S(4, 3), show that PQ and RS are
parallel.
2. Prove that the line passing through the points A(6, 4) and B(7, 11) is parallel to the line
passing through P(0, 0) and Q(2, 14).
3. Find w given that the line joining A(2, 3) to B(w, − 1) is parallel to a line with gradient −2.
C-COLLINEAR POINTS
► Three or more points are collinear if they lie on the same straight line.
If three points A, B and C are collinear, the gradient of AB is equal to the gradient of BC
and also the gradient of AC.
C
B
A
Example 3:
Prove that the points A( − 2, 5), B(1, 3) and C(7, -1) are collinear.
SOLUTION
y −y
Gradient of AB = x2 − x 1
2
3−5
(x1 , y1 ) = A( − 2, 5) and (x2 , y2 ) = B(1, 3)
1
= 1+2
−2
= 3
y −y
Gradient of BC = x2 − x1
2
1
(x1 , y1 ) = B(1, 3) and (x2 , y2 ) = C(7, − 1)
−1 − 3
= 7− 1
−4
= 6
2
= −3
Since the gradients are equal and B is common to both AB and BC, the points are collinear.
EXERCISE
1. Prove that the points A( − 2, 1), B( − 1, 0) and C(7, − 8) are collinear.
2. Show that the following points are collinear. A(1, − 1), B(6, 9) and C(3, 3).
3. Prove that the points A( − 2, 1), B( − 1, 0) and C(7, − 8) are collinear.
4. The point P has coordinates ( − 2, 4). The point Q has coordinates (6, − 4). The point R has
coordinates (3, − 1). Show that the three points are collinear.
D-Gradients of perpendicular lines
► If two lines with gradients m1 and m2 are perpendicular to each other, then the product of
their gradients is −1.
m1 × m2 = −1
1
that is, m2 = − m
2
Example 4:
If P is the point (2, − 3) and Q is the point ( − 1, 6), find the gradient of a line
perpendicular to PQ.
SOLUTION
y −y
Gradient of PQ = x2 − x1
2
1
(x1 , y1 ) = P(2, − 3) and (x2 , y2 ) = Q( − 1, 6)
6 −(−3)
= −1 − 2
9
= −3
= −3
Let m1 = −3
m1 × m2 = −1
−3 × m2 = −1
1
m2 = 3
Divide each side by −3
1
The gradient of a line perpendicular to PQ is 3.
Example 5:
Find w given that the line joining D( − 1, − 3) to C(1, w) is perpendicular to a
line with gradient 2.
SOLUTION
1
Gradient of DC = − 2 (perpendicular to the line of gradient 2)
y −y
Gradient of DC = x2 − x 1
2
1
(x1 , y1 ) = D( − 1, − 3) and (x2 , y2 ) = C(1, w)
w −(−3)
= 1 −(−1)
−1
2
=
w+3
2
Cross multiply
2(w + 3) = −2
2w + 6 = −2
2w = −8
w = −4
EXERCISE 1
1.
Find the gradient of the line that is perpendicular to the lines with these gradients.
4
2.
3.
4.
5.
1
(a) 3
(b) − 4
(d) 8
(e)
(c) −6
3
−5
(f) 5
The diagonals of the quadrilateral ABCD are A(3, 7), B( − 1, 6), C( − 2, − 3) and D(11, 0).
Show that the line AC and BD are perpendicular.
The line joining the points K( − 6, 0) and D( − 1, w) is perpendicular to a line with gradient
10. Find the value of w.
The points P, Q, R and S have coordinates ( − 5, 5), ( − 2, − 1), (1, 3) and ( − 3, 1)
respectively. Show that PQ is perpendicular to RS.
Find the gradient of each line AB. Then find the gradient of a line perpendicular to it.
(a) A(1, 4) and B(5, 0)
(b) A( − 5, − 2) and B(3, 2)
(c) A( − 1, − 2) and B(1, 4)
(d) A( − 2, − 7) and B(3, 3)
THE EQUATIONS OF STRAIGHT LINE
Vertical Lines
All the points on a vertical line must have the same x-coordinate, but the y-coordinate can take
any value.
► The vertical line through the point (x1 , y1 ) has equation x = x1 .
y
(x1 , y1 )
x1
O
x
Horizontal Lines
► The horizontal line through the point (x1 , y1 ) has equation y = y1 .
y
y1
O
(x1 , y1 )
x
Point-Gradient Form
The gradient of a line may be found if its gradient and the coordinates of any point that it
passes through are known.
► The equation of a straight line with gradient m and passing through (x1 , y1 ) is:
y − y1 = m(x − x1 )
Example 6:
Find the equation of line with gradient 4 passing through ( − 2, − 5).
SOLUTION
Using y − y1 = m(x − x1 ), where m = 4
y − (−5) = 4[x − (−2)]
y + 5 = 4(x + 2)
y + 5 = 4x + 8
y = 4x + 3
and (x1 , y1 ) = ( − 2, 5)
Example 7:
Find the equation of line passing through W( − 2, 0) and Z(1, 6).
SOLUTION
First find the gradient of the line WZ.
y −y
Gradient of WZ = x2 − x 1
(x1 , y1 ) = W( − 2, 0) and (x2 , y2 ) = Z(1, 6)
2
1
6−0
= 1 −(−2)
6
=3
=2
Using y − y1 = m(x − x1 ), where m = 2
y − 0 = 2[x − ( − 2)]
y = 2(x + 2)
y = 2x + 4
and (x1 , y1 ) = ( − 2, 0)
EXERCISE
Find the gradient of the line that passes through the points (5, 7) and (3, -1) and hence find
the equation of the line.
2. Find the equation of the line joining (2,-1) to (3,4).
3. Calculate the equation of a straight line passing through the points (−3,3) and (5,5).
4. Find the equation of the line passing through (1, 1) and (4,−8).
5. Find the equation of the line passing through (3, 4) and (5, 4).
6. Find the equation of the line passing through (0, 2) and (4, 0).
7. Find the equation of the line passing through (−2, 3) and (2,−5).
8. Find the equation of the line passing through P(−3, 4) and Q(1, 2).
9. Find the equation of the line passing through T(8, 6) and V(2, 12).
10. Find the equation of the line passing through R(−1, 1) and S(0,6).
1.
Gradient – intercept form.
► The gradient-intercept form of the linear equation is y = mx + c, where:
m is the gradient, and c is the y-intercept.
Example 8:
Write down the gradient of the line 3x − 2y − 12 = 0.
SOLUTION
Solve for the equation 3x − 2y − 12 = 0 for y.
3x − 2y − 12 = 0
3x − 12 = 2y
2y = 3x − 12
Divide each side by 2
3
y = 2x − 6
3
The gradient m = 2
Example 12: The diagram below shows a straight line passing through A(6, 0) and B(0, 9).
y
(0, 9)
(6, 0)
x
(a) Find the equation of the line AB.
(b) Find the equation of the line through A(6, 0) perpendicular to AB.
SOLUTION
y −y
(a) Gradient of AB = x2 − x1
2
0−9
1
(x1 , y1 ) = A(0, 9) and (x2 , y2 ) = C(6, 0)
= 6−0
−9
= 6
3
= −2
Use y = mx + c
3
y = −2x + 9
3
(b) m1 × m2 = −1
m1 = − 2
3
− 2 × m2 = −1
2
m2 = 3
3
m = − 2 and c = 9
3
2
multiply each side by the reciprocal of − 2 which is − 3
and (x1 , y1 ) = C(6, 0)
y − y1 = m(x − x1 )
2
y − 0 = 3 (x − 6)
2
y= x−4
3
EXERCISE
1. Find the equation of the line which has gradient 6 and crosses the y-axis at the point
(0, − 4).
2. Find the equation of the straight line with gradient of 2 and 𝑦‐intercept of −5.
3. Find the gradient and 𝑦-intercept of the line with equation 2y − 6x = 3.
4. Find the gradient and 𝑦-intercept of the line with equation 2y + 3x – 12 = 0.
5. Find the gradient and coordinates of the 𝑦-intercept of the straight lines represented by
the following equations and sketch their graphs.
(a) 3x + 3y − 7 = 0
(b) 2x − 5y + 1 = 0
(c) 5x + 10y − 2 = 8
y
6. Find the equation of the line AB.
B(0, 5)
A(−3, 0)
x
O
7. The equation of the line shown is given by 3x + y − 6 = 0. Find the gradient and
coordinates of A and B.
y
B
O
A
8. A line passes through the points A( − 6, 3) and B(6, − 5).
y
7
6
A( − 6, 3)
5
46
3
2
1
-6 -5 -4 -3 -2 -1 0
-1
1 2
3 4
5 6 7 x
-2
-3
-4
-5
-6
(a) Find the gradient of the line AB.
(b) Find the equation the line AB.
B(6, − 5)
x
EQUATIONS OF PARALLEL AND PERPENDICULAR LINES
Example 9:
(a) Find the gradient of a straight line that is parallel to the line 2y + 6x − 8 = 0.
(b) Find the gradient of a straight line that is perpendicular to the line 6x − 2y + 7 = 0.
SOLUTION
(a) Solve the equation 2y + 6x − 8 = 0 for y.
2y = −6x + 8 Divide each term by 2
y = −3x + 4
For parallel lines m1 = m2 , the gradient is −3.
(b) Solve the equation 6x − 2y + 7 = 0 for y.
−2y = −6x − 7
Divide each term by −2
7
y = −3x +
2
1
m1 = −3
m2 = − m
1
1
m2 = 3
Example 11:
Find the equation of the line passing through B(0, -2) and perpendicular to a
2
line with gradient 3.
SOLUTION
3
The perpendicular line has gradient − 2.
To find the equation use y = mx + c
3
3
where m = − 2 and c = −2
y = −2x − 2
EXERCISE
1.
2.
3.
Show that the line y = 3x + 4 is perpendicular to the line x + 3y − 3 = 0.
Show that the line 3x − y − 2 = 0 is perpendicular to the line x + 3y − 6 = 0
Work out whether the pairs of lines are parallel, perpendicular or neither.
(a) y = 5x − 3 and 5x − y + 4 = 0
(b) 4x − 5y + 10 = 0 and 8x − 10y − 2 = 0
(c) 3x + 2y − 12 = 0 and 2x + 3y − 6 = 0 (d) 5x − y + 2 = 0 and 2x + 10y − 4 = 0
4.
5.
Find the equation of the line parallel to the line y = 3x − 11 and passes through point
( − 5, 9).
Find the equation of the straight line that passes through the point (4, 6) and is parallel to
the line y = 3x − 8.
Find the equations of the lines parallel to y = −3x + 1 that pass through point (−4,1).
Find the equation of the line parallel to the line y = 4 − 3x and passing through the point
(5, -3).
8. Find the equation of the line through the point ( − 1, 3) parallel to the line with equation
2x − 5y = 10.
9. Find the equation of the line perpendicular to the line 3y = 15x and passing through the
point (4, − 3).
10. Find the equation of the line through the point ( − 1, 3) perpendicular to the line with
equation 6x + 9y − 12 = 0.
11. Find the equation of the line through the point (2, − 5) perpendicular to the line with
equation 6x − 3y + 15 = 0.
6.
7.
Midpoint
► The midpoint of a straight line with end points A(x1 , y1 ) and Q(x2 , y2 ) is:
x +x
y +y
M ( 1 2 2 , 1 2 2)
Take the average of the x-coordinate and the average of the y-coordinate.
Example 11:
Find the coordinates of the midpoint of the line segment AB, given:
(a) A(1, 2) and B(7, 2)
(b) A( − 5, − 2) and B( − 7, 4 )
SOLUTION
(a) Let (x1 , y1 ) = A(1, 2) and (x2 , y2 ) = B(7, 2)
1+7 2+2
The midpoint has coordinates ( 2 , 2 ) = (4, 2)
(b) Let (x1 , y1 ) = A( − 5, − 2) and (x2 , y2 ) = B( − 7, 4)
−5+(−7) −2+4
The midpoint has coordinates (
Example 12:
SOLUTION
Let the coordinates of H be (x1 , y1 ),
=3
x1 − 2 = 6
x1 = −8
y1 + 2 = 12
y1 = 10
2
,
2
) = ( − 6, 1)
If M(3, 6) is the midpoint of the straigth line HK and K has coordinates
( − 2, 2), find the coordinates of H.
x1 +(−2)
2
and
y1 +2
2
=6
Thus H has coordinates ( − 8, 10)
Exercise
1.
2.
3.
4.
5.
Find the coordinates of the midpoint of the line segment AB where A and B have
coordinates.
(a) A(8, 0) and B(4, 6)
(b) A( − 6, − 2) and B( − 4, 6)
(c) A( − 2, − 8) and B(5, − 8)
(d) A(2w, x) and B( − 6w, 7x)
(e) A(8x, 2y) and B(2x, − 14y)
(f) A( − 2, 6) and B( − 2, − 14)
The coordinates of points R and S are ( − 4, 3) and (x, y) respectively. Point M( − 1, 2)
is the midpoint of RS. Find the values of x and y.
Point P has the coordinates P(1, 8). The midpoint, M, of the line segment AB has the
coordinates M(5, 3). Find the coordinates of B.
The coordinates of the endpoint, C, and the midpoint, M, of the line segment CD are
(6, − 7) and M(2, − 1). Find the coordinates of point D.
The midpoint M of the line segment AB has coordinates ( − 3, 1). B has coordinates
( − 1, − 2). Find the coordinates of the point A.
THE DISTANCE BETWEEN TWO POINTS
Distance is a measure of the length between two points.
Apply the theorem of Pythagoras in triangle ABC to find the length of AB.
y
4
3
B
2
1
-6 -5 -4 -3 -2 -1 0
-1
A
AB2 = AC2 + BC2
-2
-3
1 2
3 4 5 x
C
∴AB = √AC2 + BC2
AC = 4 − ( − 4)
=8
BC = 4 − ( − 2)
=6
AB2 = 82 + 62
AB2 = 64 + 36
AB2 = 100
AB = √100
AB = 10 units
Given that the points ABC have coordinates A(x1 ,y1 ), B(x2 ,y2 ) and C(x2 ,y1 ):
AC = x2 − x1
BC = y2 − y1
AB = √(x2 − x1 )2 + (y2 − y1 )
2
Therefore to calculate the distance between two points A(x1 ,y1 ) and B(x2 ,y2 ),
we use Distance d = √(x2 − x1 )2 + (y2 − y1 )
2
Example: Find the distance between M(7, 13) and N(2, 1).
SOLUTION
d = √(x2 − x1 )2 + (y2 − y1 )
2
(x1 , y1 ) = M(7, 13) and (x2 , y2 ) = N(2, 1)
d = √(2 − 7)2 + (1 − 13)2
= √(−5)2 +(−12)2
= √25 + 144
= √169
= 13 units
EXERCISE
1.
2.
3.
4.
Find the distance between P( − 2, − 5) and R(7, − 2).
Given the points A(−5, 2), B(4, 5) and C(4, 2). How do we find the length of AB?
Given that the distance between KL = 13, R(3, 9) and S(8,y). Find y.
If the distance between C(0,-3) and F(8,w) is 10 units, find the possible values of w.
5.
6.
Find the distance between each of the following points:
(a) P(6, 8) and Q( − 9, − 12)
(b) A( − 6, − 1) and B( − 6, − 11)
The length of CD = 5. Find the missing coordinate if
(a) C(6, − 2) and D(x, 2)
(b) C(4,y) and D(1, − 1)