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 ) P M : P N = 2 : 3 P M P N = 2 3 3 P M = 2 P N 3 ( x ( 3 ) ) 2 + ( y 5 ) 2 = 2 ( x 4 ) 2 + ( y 7 ) 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
( y 2 x 0 ) ( y 5 x 6 ) = 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.

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