Coordinate Geometry Short Questions (Question 6 & 7)


Question 6:
The point M is (–3, 5) and the point N is (4, 7). The point P moves such that PM: PN = 2: 3. Find the equation of the locus of P.

Solution:
Let P=( x,y ) PM:PN=2:3 PM PN = 2 3 3PM=2PN 3 ( x( 3 ) ) 2 + ( y5 ) 2 =2 ( x4 ) 2 + ( y7 ) 2
Square both sides to eliminate the square roots.
9[x2 + 6x + 9 + y2 – 10y + 25] = 4 [x2 – 8x + 16 + y2 – 14y + 49]
9x2 + 54x + 9y2 – 90y + 306 = 4x2 – 32x + 4y2 – 56y + 260
5x2 + 5y2 + 86x – 34y + 46 = 0

Hence, the equation of the locus of point P is
5x2 + 5y2 + 86x – 34y + 46 = 0



Question 7:
Given the points A (0, 2) and B (6, 5). Find the equation of the locus of a moving point P such that the triangle APB always has a right angle at P.
Solution:
Let P = (x, y)
Given that triangle APB90o, thus AP is perpendicular to PB.
Hence, (mAP)(mPB) = –1.

(mAP)(mPB) = –1
( y2 x0 )( y5 x6 )=1
(y – 2)(y – 5) = – x(x – 6)
y2 – 7y + 10 = –x2 + 6x
y2 + x2 – 6x – 7y + 10 = 0

Hence, the equation of the locus of point P is
y2 + x2 – 6x – 7y + 10 = 0.