The point moves such that the area of the tri | vedclass

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Question

(A) Let the point be $(x, y)$.The area of the triangle with vertices $(x, y)$,$(1, 5)$,and $(3, -7)$ is given by the formula:$\frac{1}{2} |x(5 - (-7))

Vedclass · Accepted Answer

$6x + y - 32 = 0$

Vedclass · Answer

(A) Let the point be $(x, y)$.The area of the triangle with vertices $(x, y)$,$(1, 5)$,and $(3, -7)$ is given by the formula:$\frac{1}{2} |x(5 - (-7)) + 1(-7 - y) + 3(y - 5)| = 21$$\frac{1}{2} |x(12) - 7 - y + 3y - 15| = 21$$\frac{1}{2} |12x + 2y - 22| = 21$$|6x + y - 11| = 21$This gives two possible equations:$6x + y - 11 = 21 \implies 6x + y - 32 = 0$$6x + y - 11 = -21 \implies 6x + y + 10 = 0$Comparing with the given options,the locus is $6x + y - 32 = 0$.

The point moves such that the area of the triangle formed by it with the points $(1, 5)$ and $(3, -7)$ is $21$ sq. units. The locus of the point is:

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