The point (-4, 5) is a vertex of a square and | vedclass

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(C) Let the given vertex be $V = (-4, 5)$ and the given diagonal be $L_1: 7x - y + 8 = 0$. Note that the vertex $V(-4, 5)$ does not lie on the diagona

Vedclass · Accepted Answer

$7y + x = 31$

Vedclass · Answer

(C) Let the given vertex be $V = (-4, 5)$ and the given diagonal be $L_1: 7x - y + 8 = 0$. Note that the vertex $V(-4, 5)$ does not lie on the diagonal $L_1$ because $7(-4) - 5 + 8 = -28 - 5 + 8 = -25 
eq 0$. In a square, the diagonals are perpendicular to each other and bisect each other. Let the other diagonal be $L_2$. Since $L_2$ passes through the vertex $V(-4, 5)$ and is perpendicular to $L_1$, its slope must be the negative reciprocal of the slope of $L_1$. The slope of $L_1$ is $m_1 = 7$. Therefore, the slope of $L_2$ is $m_2 = -1/7$. The equation of $L_2$ passing through $(-4, 5)$ is: $y - 5 = -\frac{1}{7}(x + 4)$ $7(y - 5) = -(x + 4)$ $7y - 35 = -x - 4$ $x + 7y = 31$ Thus, the equation of the other diagonal is $x + 7y = 31$ or $7y + x = 31$.

The point $(-4, 5)$ is a vertex of a square and one of its diagonals lies along the line $7x - y + 8 = 0$. The equation of the other diagonal is:

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