The locus of a point which moves such that th | vedclass

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(D) Let the point be $P(x, y)$. The vertices of the triangle are $A(1, 2)$,$B(-2, 5)$,and $P(x, y)$.The area of the triangle is given by the formula:

Vedclass · Accepted Answer

$3x + 3y + 7 = 0 \quad \& \quad 3x + 3y - 25 = 0$

Vedclass · Answer

(D) Let the point be $P(x, y)$. The vertices of the triangle are $A(1, 2)$,$B(-2, 5)$,and $P(x, y)$.The area of the triangle is given by the formula: $	ext{Area} = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$.Substituting the values: $8 = \frac{1}{2} |1(5 - y) + (-2)(y - 2) + x(2 - 5)|$.$16 = |5 - y - 2y + 4 - 3x| = |9 - 3y - 3x|$.$16 = |9 - 3(x + y)|$.This gives two cases:Case $1$: $9 - 3(x + y) = 16 \implies -3(x + y) = 7 \implies 3x + 3y + 7 = 0$.Case $2$: $9 - 3(x + y) = -16 \implies -3(x + y) = -25 \implies 3x + 3y - 25 = 0$.Thus,the locus is $3x + 3y + 7 = 0$ and $3x + 3y - 25 = 0$.

The locus of a point which moves such that the area of the triangle formed by it with the vertices $(1, 2)$ and $(-2, 5)$ is $8$ sq. units is/are

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