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(c) Obviously, slope of $AC = 5/2$.

Let m be the slope of a line inclined at an angle of ${45^o}$ to $AC$, then $	an {45^o} = \pm \frac{{m - \frac

Vedclass · Accepted Answer

$D\,\left( {\frac{9}{2},\,\,\frac{1}{2}} \right)\,,\,\,B\,\left( { - \frac{1}{2},\,\,\frac{5}{2}} \right)$

Vedclass · Answer

(c) Obviously, slope of $AC = 5/2$.

Let m be the slope of a line inclined at an angle of ${45^o}$ to $AC$, then $	an {45^o} = \pm \frac{{m - \frac{5}{2}}}{{1 + m.\frac{5}{2}}} \Rightarrow m = - \frac{7}{3},\frac{3}{7}$.

Thus, let the slope of $AB$ or $DC$ be $3/7$ and that of $AD$ or $BC$ be $ - \frac{7}{3}$ . Then equation of $AB$ is $3x - 7y + 19 = 0$.

Also the equation of $BC$ is $7x + 3y - 4 = 0$.

On solving these equations, we get, $B\,\,\,\left( { - \frac{1}{2},\frac{5}{2}} ight)$.

Now let the coordinates of the vertex $D$ be $(h, k)$. Since the middle points of $AC$ and $BD$ are same,

therefore $\frac{1}{2}\left( {h - \frac{1}{2}} ight)\, = \frac{1}{2}(3 + 1) \Rightarrow h = \frac{9}{2}$, $\frac{1}{2}\left( {k + \frac{5}{2}} ight) = \frac{1}{2}(4 - 1)$

==> $k = \frac{1}{2}$. Hence, $D = \left( {\frac{9}{2},\,\frac{1}{2}} ight)$.

The opposite angular points of a square are $(3,\;4)$ and $(1,\; - \;1)$. Then the co-ordinates of other two points are

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