Question

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

• A. $(2,3)$
• B. $(3,2)$
• C. $(3,3)$
• D. $(2,2)$

Answer 1

The correct option is C $(3,3)$

Let $A(4,0)$ & $B(0,2)$ be the intersection point of line $x+2y=4$ on X-axis and Y-axis respectively.

Now let $ABCD$ be the square formed keeping AB as base on line, away from origin.

We know length $l\left(AB\right)=\sqrt{4^2+2^2}=2\sqrt{5}$

So, Side of square= $2\sqrt{5}$

Now to find coordinates of $D$, we know $\angle BAO=ta{n}^{-1}\frac{1}{2}$

so $\angle DAE=\pi -(\frac{\pi }{2}+ta{n}^{-1}\frac{1}{2})\Rightarrow ta{n}^{-1}\left(2\right)$.

We know $l\left(AD\right)=2\sqrt{5}$ and $\angle DAE=ta{n}^{-1}\left(2\right)\Rightarrow co{s}^{-1}\left(\frac{1}{\sqrt{5}}\right)=si{n}^{-1}\left(\frac{2}{\sqrt{5}}\right)$

so, $l\left(AE\right)=2\sqrt{5}cos\left(co{s}^{-1}\right(\frac{1}{\sqrt{5}}\left)\right)\Rightarrow 2$

Hence we can find $X-coords$. of $D=l\left(OA\right)+l\left(AE\right)=6$

Now, we know $\angle DAE=ta{n}^{-1}\left(2\right)$ & $l\left(AE\right)=2$ , so

$l\left(DE\right)=2tan\left(ta{n}^{-1}\right(2\left)\right)=4$.

Hence, we get $Y-coordinate$ of $D=l\left(DE\right)\Rightarrow 4$

$\therefore$ $D=(6,4);B=(0,2)$

As we know that Diagonals of a square bisect each other. So, Coordinates of Intersection of diagonals is Midpoint of Diagonal $BD$ i.e $(\frac{6+0}{2},\frac{4+2}{2})\Rightarrow (3,3)$

Hence, Option $\left(C\right)$ is correct.