 # 6.7.3 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:

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
$\left(\frac{y-2}{x-0}\right)\left(\frac{y-5}{x-6}\right)=-1$
(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.