Question

On the portion of the straight line,$\text{x+2y = 4}$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

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

Answer 1

The correct option is C

$3,3$

Step 1: Finding the co-ordinates of D:

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

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

We know length $l\left(AB\right)=$$\sqrt{4^2+2^2}=\sqrt{20}=2\sqrt{5}$​

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

Now to find the coordinates of $D$, we know ${\text{∠BAO = tan}}^{-1}\frac{1}{2}$

​so$\text{∠DAE = π - (}\frac{\mathrm{\pi }}{2}{\text{+tan}}^{-1}\frac{1}{2}{\text{) = tan}}^{-1}\text{2}$

We know $\text{length of AD = 2}\sqrt{5}{\text{and ∠DAE = tan}}^{-1}{\text{2 = cos}}^{-1}\frac{1}{\sqrt{5}}{\text{= sin}}^{-1}\frac{2}{\sqrt{5}}$

so,$\text{length of AE = 2}\sqrt{5}{\text{cos(cos}}^{-1}\frac{1}{\sqrt{5}}\text{) = 2}\sqrt{5}\text{×}\frac{1}{\sqrt{5}}\text{= 2}$

Hence we can find $\text{X-coordinates of D = length of OA + length of AE = 4+2 = 6}$

Now, we know${\text{∠DAE = tan}}^{-1}\text{2 and length of AE = 2}$,

so ${\text{length of DE = 2 tan(tan}}^{-1}\text{2) = 2×2 = 4}$

Hence, we get $\text{Y-coordinate of D = length of ED = 4}$

$\text{Therefore, D = (6, 4), B = (0,2)}$

Step 2: Finding the mid-point of $\text{BD}$:

As we know that Diagonals of a square bisect each other.

So, the coordinates of the intersection of diagonals is the Midpoint of Diagonal $\text{BD}$ $\text{i.e. (}\frac{6+0}{2}\text{,}\frac{2+4}{2}\text{) = (}\frac{6}{2}\text{,}\frac{6}{2}\text{) = (3, 3)}$

Hence, the correct answer is $(3,3)$.