One diagonal of a square is along the line 8x | vedclass

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(C) Let the diagonal be $BD$ along the line $8x - 15y = 0$. The slope of $BD$ is $m_1 = \frac{8}{15}$.Let the vertex be $D(1, 2)$. The sides passing t

Vedclass · Accepted Answer

$23x - 7y - 9 = 0, 7x + 23y - 53 = 0$

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(C) Let the diagonal be $BD$ along the line $8x - 15y = 0$. The slope of $BD$ is $m_1 = \frac{8}{15}$.Let the vertex be $D(1, 2)$. The sides passing through $D$ are $AD$ and $CD$.In a square,the diagonal makes an angle of $45^\circ$ with the sides.Let the slope of the sides be $m$. Then,$	an(45^\circ) = \left| \frac{m - \frac{8}{15}}{1 + m \cdot \frac{8}{15}} ight|$.$1 = \left| \frac{15m - 8}{15 + 8m} ight|$.This gives two cases:Case $1$: $15m - 8 = 15 + 8m$ $\Rightarrow 7m = 23$ $\Rightarrow m = \frac{23}{7}$.Case $2$: $15m - 8 = -(15 + 8m)$ $\Rightarrow 15m - 8 = -15 - 8m$ $\Rightarrow 23m = -7$ $\Rightarrow m = -\frac{7}{23}$.The equations of the sides passing through $(1, 2)$ are:$y - 2 = \frac{23}{7}(x - 1)$ $\Rightarrow 7y - 14 = 23x - 23$ $\Rightarrow 23x - 7y - 9 = 0$.$y - 2 = -\frac{7}{23}(x - 1)$ $\Rightarrow 23y - 46 = -7x + 7$ $\Rightarrow 7x + 23y - 53 = 0$.

One diagonal of a square is along the line $8x - 15y = 0$ and one of its vertices is $(1, 2)$. Then the equations of the sides of the square passing through this vertex are:

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