P(x, y) moves such that the area of the trian | vedclass

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Question

(A) The area of a triangle with vertices $(x_1, y_1), (x_2, y_2), (x_3, y_3)$ is given by $\frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_

Vedclass · Accepted Answer

$3x - y = a$

Vedclass · Answer

(A) The area of a triangle with vertices $(x_1, y_1), (x_2, y_2), (x_3, y_3)$ is given by $\frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$.For $\Delta PQR$ with $P(x, y), Q(a, 2a), R(-a, -2a)$:Area $= \frac{1}{2} |x(2a - (-2a)) + a(-2a - y) + (-a)(y - 2a)| = \frac{1}{2} |4ax - 2a^2 - ay - ay + 2a^2| = \frac{1}{2} |4ax - 2ay| = |2ax - ay|$.For $\Delta PST$ with $P(x, y), S(a, 2a), T(2a, 3a)$:Area $= \frac{1}{2} |x(2a - 3a) + a(3a - y) + 2a(y - 2a)| = \frac{1}{2} |-ax + 3a^2 - ay + 2ay - 4a^2| = \frac{1}{2} |-ax + ay - a^2| = \frac{1}{2} |ay - ax - a^2|$.Equating the areas: $|2ax - ay| = \frac{1}{2} |ay - ax - a^2|$.Case $1$: $2ax - ay = \frac{1}{2}(ay - ax - a^2) \implies 4ax - 2ay = ay - ax - a^2 \implies 5ax - 3ay = -a^2 \implies 5x - 3y = -a$.Case $2$: $2ax - ay = -\frac{1}{2}(ay - ax - a^2) \implies 4ax - 2ay = -ay + ax + a^2 \implies 3ax - ay = a^2 \implies 3x - y = a$ (assuming $a 
eq 0$).Comparing with options,$3x - y = a$ is the correct locus.

$P(x, y)$ moves such that the area of the triangle formed by $P, Q(a, 2a)$ and $R(-a, -2a)$ is equal to the area of the triangle formed by $P, S(a, 2a)$ and $T(2a, 3a)$. The locus of $P$ is a straight line given by:

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