The point on the line 3x + y + 4 = 0 which is | vedclass

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(D) Let the point be $P(x, y)$. Since $P$ lies on the line $3x + y + 4 = 0$,we have $y = -3x - 4$. Let $A = (-5, 6)$ and $B = (3, 2)$. Since $P$ is eq

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$(-2, 2)$

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(D) Let the point be $P(x, y)$. Since $P$ lies on the line $3x + y + 4 = 0$,we have $y = -3x - 4$. Let $A = (-5, 6)$ and $B = (3, 2)$. Since $P$ is equidistant from $A$ and $B$,$PA^2 = PB^2$. $(x + 5)^2 + (y - 6)^2 = (x - 3)^2 + (y - 2)^2$. $x^2 + 10x + 25 + y^2 - 12y + 36 = x^2 - 6x + 9 + y^2 - 4y + 4$. $16x - 8y + 48 = 0$,which simplifies to $2x - y + 6 = 0$. Substitute $y = -3x - 4$ into $2x - y + 6 = 0$: $2x - (-3x - 4) + 6 = 0$. $5x + 10 = 0 \implies x = -2$. Then $y = -3(-2) - 4 = 6 - 4 = 2$. The point is $(-2, 2)$.

The point on the line $3x + y + 4 = 0$ which is equidistant from $(-5, 6)$ and $(3, 2)$ is

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