One diagonal of a square is along line 8x - 15y = 0 and one of its vertex is (1, 2) Then equation of

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9.Straight Line

hard

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

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

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

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

None of these

(IIT-1962)

Solution

(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 $\tan {45^o} = \pm \frac{{m – \frac{8}{{15}}}}{{1 + \frac{8}{{15}}m}}$

$ \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$.

Standard 11

Mathematics

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normal

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

Solution

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