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Q. On the portion of the straight line $x+2 y=4$ intercepted between the axes, a square is constructed on the side of the line away from the origin. Then the point of intersection of its diagonals has co-ordinates :

Straight Lines

Solution:

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$\tan 45^{\circ}=\left|\frac{m+\frac{1}{2}}{1-\frac{m}{2}}\right| $
$ \Rightarrow \pm 1=\frac{2 m+1}{2-m} $
$ \Rightarrow m=\frac{1}{3},-3$
$\therefore $ Equation of $AC$
$y-2=\frac{1}{3}(x) \Rightarrow x-3 y+6=0 .....$(i)
Equation of $BD$
$y=-3(x-4) \Rightarrow 3 x+y-12=0 ......$(ii)
From (i) \& (ii)
$x=3 \& y=3$