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

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(c) Slope of $BD$ is $\frac{8}{{15}}$ and angle made by $BD$ with $AD$ and $DC$ is ${45^o}$. So let slope of $DC$ be m, then $	an {45^o} = \pm \frac{

Vedclass · Accepted Answer

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

Vedclass · Answer

(c) Slope of $BD$ is $\frac{8}{{15}}$ and angle made by $BD$ with $AD$ and $DC$ is ${45^o}$. So let slope of $DC$ be m, then $	an {45^o} = \pm \frac{{m - \frac{8}{{15}}}}{{1 + \frac{8}{{15}}m}}$

$ \Rightarrow (15 + 8m) = \pm (15m - 8)$

==> $m = \frac{{23}}{7}$and $ - \frac{7}{{23}}$

Hence the equations of $DC$ and $AD$ are

$y - 2 = \frac{{23}}{7}(x - 1)$$ \Rightarrow 23x - 7y - 9 = 0$

and $y - 2 = - \frac{7}{{23}}(x - 1)$$ \Rightarrow 7x + 23y - 53 = 0$.

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

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