\

7.8.3 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)+3h2+3,2(12)+3k2+3)(0,6)=(12+3h5,24+3k5)12+3h5=0      3h=−12 h=−424+3k5=63k=30−24k=2P=(−4,2)

(a)(ii) 
mPS=2−(−6)−4−2 =−86 =−43Equation of PS:y−y1=−43(x−2)y−(−6)=−43x+833y+18=−4x+83y=−4x−10


(a)(iii) 
Area of △ PRS=12|−4   2    6  2  −6 12  −42|=12|(24+24+12)−(4−36−48)|=12|60−(−80)|=70 unit2

(b) 
Let P=(x,y)MR=2MS(x−6)2+(y−12)2=2(x−2)2+(y+6)2(x−6)2+(y−12)2=4[(x−2)2+(y+6)2]x2−12x+36+y2−24y+144=4[x2−4x+4+y2+12y+36]x2−12x+y2−24y+180=4x2−16x+4y2+48y+1603x2+3y2−4x+72y−20=0

Leave a Comment