CO-ORDINATE GEOMETRY
1. Find the length of AB if A (2,3) and B(4,5)
Y
B (4,5)
5
A(2,3
3
)
X
0 2 4
A(2,3) and B(4,5)
Using Pythagoras theorem
AQ =2 units
QB = 2 units
Therefore, AB2 = (QA)2 = (BQ)2
AB2 = (4-2)2 + (5-3)2
AB2 = (2)2 + (2)2
AB2 = 4+4
AB2 = 8
AB = √8
From above, we can say that the distance along the x-axis = (x2 – x1) and distance along
the y-axis = (y2 – y1)
AB2 = (X2 – X1)2 + (Y2 –Y1)2
AB2 = (4-1)2 + (5-3)2
AB2 = (2)2 + (2)2
AB2 = 4+4
AB = √8
Therefore, distance between two points = � (𝑋𝑋2 – 𝑋𝑋1)2 + (𝑌𝑌2 – 𝑌𝑌1)2
2. Find the midpoint of A and B
X1+ X2 , Y1 +Y2 (halfway x1 and x2 and halfway y1 and y2)
2 2
x1, y1 x2, y2
A (2,3) and B (4,5)
= 2+4, 3+5
2 2
6 8
Midpoint= � + �
2 2
Midpoint = ( 3,4 )
3. Calculate the gradient of the line C(-5,-3) and D(-2,6) and E(1,2) and (3,-1)
.D (-2,6) 6
2 .E(1,2)
-5 -4 -3 -2 -1 0 1 2 3
-1 . F (3, -1)
-2
. C (-5,-3) -3
M = ∆y = M = Y1-Y2 C = (-5, -3) and D (-2, 6)
∆x x1-x2
MCD = 6-(-3)
-2-(-5)
MCD = 6+3
-2+5
MCD = 9
3
MCD = 3
TIP - since the line slops to the right, the gradient should be positive
X1,Y1 , X2 , Y2
(ii) MEF =∆ Y = MEF = Y2 -Y1 EF = (1,2) and (3,-1)
∆X X2 –X1
MEF = -1-2
3-1
MEF = -3
2
Since the line is sloping to the left, the gradient should be negative
4. (i) Find the equation of the straight line through (-2,-3) with a gradient 2.
Y – Y1 = M (X – X1) (equation to use when a point on the line
Y-(-3)=2(X-(-2)) is given and gradient) M=Y2-Y1
Y+3=2 (X+2) X2-X1
Y+3=2X +4
Y=2X +4 -3
Y=2X+1
(ii) Find the equation of the straight line through (2,4) and (-2, -4)
M = Y-Y1
X-X1
Y-Y1 = M (X-X1) Since we do not have M then we find M as y2 –y1
Y – Y1 = y2 –y1 x2 –x1
x2 –x1
Y -4 = -4-4
-2-2 (X-2) using the point (2,4) to substitute in to the equation
Y – 4= 2 (X-2)
Y=2X-4+4
Y = 2X
(iii) Find the equation of the straight line through the point (4,-2) and the origin
i.e (4,-2) and (0,0) origin has co-ordinates(0,0)
Y- Y1 = M (X-X1)
Y-Y1 = Y2-Y1 ,(X-X1)
X2-X1
Y-0 = 0-(-2) (X-0) using the point (0,0)
0-4
Y = -2X OR 4Y = -2X
4
(iv) Find the equation of the straight line with the gradient 3 and y – intercept of 6
Y = MX +C , M is the gradient, C is the y-intercept, the point
Where the line cuts the y-axis
Therefore, Y = 3X + 6 By substituting the values of M and C
(V) Find the equation of the straight line with the gradient -2/9 and x-intercept 3
Y = MX+C X-intercept is the point where the line cuts the x-axis. Along
−2
Y= (3) +C x-axis all the y- co-ordinates are zero(0)
9
−2
Y= +C therefore, the point is (3,0)
3
−2
0= +C
3
2
C=
3
The equation of the straight line is
−2 2
y= x+ or 9y = -2x +6 ,by multiplying through by 9.
9 3
(vi) Find the equation of the straight line with an x- intercept 4 and y- intercept 6.
i.e (4,0) and (0,6)
M = Y2+Y1
X2+Y1
M = 6-0
0-4
−6 −3
M= =
4 2
Using the point (0,6) i.e C=6
Y = MX +C
−3
Y= x+6. Or
2
2y = -3x = 12 multiplying throughout by 2.
5.(i) Find the equation of the straight line that passes through the point (3,8) and is
parallel to y = 2x -9
- The gradient of the new line is 2 since parallel lines have the same gradient
i.e M1 = M2.
Using the point (3,8)
Y = MX +C
8 = 2(3) + C
8 = 6 +C
2=C
Y = 2X +2
(ii) Find the equation of the line that is perpendicular to 3y – 2x = 4 and passes through
the point (-7,4)
3y = 2x +4 making y the subject of
y = 2x + 4 the formula
3 3
Therefore, M = 2
3
- Gradient of parallel lines M2× M2 = -1
2�
3 × 𝑀𝑀2 = −1
2M2 = -3
−3
M2 =
2
Using the point (-7,4) to substitute into the equation
Y = MX + C,
−3 (−7)
4= +C
2
4 = 21 +C
2
4 -21 = C
1 2
8-21 = C
2
-13 =C
2
The equation perpendicular to 3y-2x=4 which passes through point (-7,4) is
−3𝑥𝑥 13
Y= - or
2 2
2y = -3x - 13
PRACTICE QUESTIONS:
(1) Find the gradient of the straight line whose equation is 3y + x = 5.
(2) Find the equation of the straight line passing through (-4, 4) and is
𝑥𝑥
perpendicular to the straight line whose equation is 𝑦𝑦 + = 1.
7
(3) In the diagram below, the points A and B are (4,0) and (0,8) respectively.
Find the equation of AB.
(4) In the diagram, B is the point (0,16) and C is the point (0,6). The sloping line
through B and the horizontal line through C meet at the point A.
(a) Write down the equation of the line AC.
(b) Given that the gradient of the line AB is 2, find the equation of the line AB.
(c) Calculate the coordinates of the point A.
(d) Calculate the area of the triangle ABC.
ANSWERS:
1
(1) m = −
3
(2) y = 7x + 32
(3) y = -2x + 8
(4) (a) y = 6
(b) y = 2x + 16
(c ) A(-5, 6)
(d ) area = 25 units2
