Coordinate Geometry Long Questions (Question 3 & 4)


Question 3:
In the diagram, PRS and QRT are straight lines. Given R is the midpoint of PS and
QR : RT = 1 : 3, Find
(a) the coordinates of R,
(b) the coordinates of T,
(c) the coordinates of the point of intersection between lines PQ and ST produced.

Solution:
(a)
                                                   Given R is the midpoint of PS.
R=( 3+7 2 , 2+6 2 ) R=( 5, 4 )                                                 
(b)
QR:RT=1:3 Lets coordinates of T=(x, y) ( ( 1 )( x )+( 3 )( 4 ) 1+3 , ( 1 )( y )+( 3 )( 5 ) 1+3 )=( 5, 4 ) x+12 4 =5 x+12=20 x=8 y+15 4 =4 y+15=16 y=1 T=(8, 1)
(c)
Gradient of PQ= 52 43 =3 Equation of PQ, y2=3( x3 ) y2=3x9 y=3x7(1) Gradient of ST= 61 78 =5 Equation of ST, y1=5( x8 ) y1=5x+40 y=5x+41(2)
                                                 Substitute (1) into (2),
                                                 3x – 7 = –5x + 41
                                                 8x = 48
                                                 x = 6

                                                 From (1),
                                                 y = 3(6) – 7 = 11 

The coordinates of the point of intersection between lines PQ and ST = (6, 11).



Question 4:

The diagram shows a triangle LMN where L is on the y-axis. The equation of the straight line LKN and MK are 2y – 3x + 6 = 0 and 3y + x – 13 = 0 respectively. Find
(a) the coordinates of K
(b) the ratio LK:KN

Solution:
(a)
2y – 3x + 6 = 0 ----(1)
3y + x – 13 = 0 ----(2)
x = 13 – 3y ----(3)

Substitute equation (3) into (1),
2y – 3 (13 – 3y) + 6 = 0
2y – 39 + 9y + 6 = 0
11y = 33
y = 3
Substitute y = 3 into equation (3),
x = 13 – 3 (3)
x = 4
Coordinates of K = (4, 3).

(b)
Given equation of LKN is 2y – 3x + 6 = 0
At y – axis, x = 0,
x coordinates of point L = 0.
Ratio LK:KN Equating the x coordinates, LK(10)+KN(0) LK+KN =4 10LK=4LK+4KN 6LK=4KN LK KN = 4 6 LK KN = 2 3 Ratio LK:KN=2:3