Short Questions (Question 6 & 7)

Posted on April 21, 2020 by user

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:

\begin{array}{l} Let P = (x, y) \\ P M : P N = 2 : 3 \\ \frac{P M}{P N} = \frac{2}{3} \\ 3 P M = 2 P N \\ 3 \sqrt{{(x - (- 3))}^{2} + {(y - 5)}^{2}} = 2 \sqrt{{(x - 4)}^{2} + {(y - 7)}^{2}} \end{array}

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 APB= 90^o, thus AP is perpendicular to PB.

Hence, (m_AP)(m_PB) = –1.

(m_AP)(m_PB) = –1

(\frac{y - 2}{x - 0}) (\frac{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.