Coordinate Geometry Long Questions (Question 5 & 6)


Question 5:
In the diagram, the equation of FMG is y = – 4. A point P moves such that its distance from E is always half of the distance of E from the straight line FG. Find
(a) The equation of the locus of P,
(b) The x-coordinate of the point of intersection of the locus and the x-axis.

Solution:
(a) 
Gradient of the straight line FMG = 0
EM is perpendicular to FMG, so gradient of EM also = 0, equation of EM is x = 2
Thus, coordinates of point M = (2, 4).

Let coordinates of point P= (x, y).
Given PE = ½ EM
2PE = EM
2 [(x – 2)2+ (y – 4)2]½ = [(2 2)2 + (4 (4))2]½
4 (x2 – 4x + 4 + y2 – 8y +16) = (0 + 64) → (square for both sides)
4x2 – 16x + 16 + 4y2 – 32y + 64 = 64
4x2 + 4y2 – 16x – 32y + 16 = 0
x2 + y2 – 4x – 8y + 4 = 0

(b) 
x2 + y2 – 4x – 8y + 4 = 0
At x axis, y = 0.
x2 + 0 – 4x – 8(0) + 4 = 0
x2  – 4x+ 4 = 0
(x – 2) (x – 2) = 0
x = 2

The x-coordinate of the point of intersection of the locus and the x-axis is 2.


Question 6:
Solutions by scale drawing will not be accepted.
Diagram below shows a triangle PRS. Side PR intersects the y-axis at point Q.

(a) Given PQ : QR = 2 : 3, find
(i) The coordinates of P,
(ii) The equation of the straight line PS,
(iii) The area, in unit2, of triangle PRS.
(b) Point M moves such that its distance from point R is always twice its distance from point S.
Find the equation of the locus M.

Solution:
(a)(i) 
P=( 2( 6 )+3h 2+3 , 2( 12 )+3k 2+3 ) ( 0,6 )=( 12+3h 5 , 24+3k 5 ) 12+3h 5 =0        3h=12  h=4 24+3k 5 =6 3k=3024 k=2 P=( 4,2 )

(a)(ii) 
m PS = 2( 6 ) 42  = 8 6  = 4 3 Equation of PS: y y 1 = 4 3 ( x2 ) y( 6 )= 4 3 x+ 8 3 3y+18=4x+8 3y=4x10

(a)(iii) 
Area of  PRS = 1 2 | 4   2    6   2  6  12   4 2 | = 1 2 | ( 24+24+12 ) ( 43648 )| = 1 2 | 60 ( 80 )| =70  unit 2

(b) 
Let P=( x,y ) MR=2MS ( x6 ) 2 + ( y12 ) 2 =2 ( x2 ) 2 + ( y+6 ) 2 ( x6 ) 2 + ( y12 ) 2 =4[ ( x2 ) 2 + ( y+6 ) 2 ] x 2 12x+36+ y 2 24y+144=4[ x 2 4x+4+ y 2 +12y+36 ] x 2 12x+ y 2 24y+180=4 x 2 16x+4 y 2 +48y+160 3 x 2 +3 y 2 4x+72y20=0

Leave a Comment