Question

$A(1,3)$ and $C(7,5)$ are two opposite vertices of a square. Then the equation of a side through $A$ is

• A. $x+2y-7=0$
• B. $x-2y+5=0$
• C. $2x+y-5=0$
• D. $2x-y+1=0$

Answer 1

The correct option is A $x+2y-7=0$

Let $ABCD$ be a square, whose diagonal is $AC$

Now, the angle between $AC$ and side $AB$ is ${45}^{\circ }$.

Let the equation of the side through $A$ is $y=mx+c$ ....... $\left(1\right)$

Now, equation of $AC$ is

$\frac{y-5}{3-5}=\frac{x-7}{1-7}$

$3y-15=x-7$

$3y=x+8$ ....... $\left(2\right)$

Slope of line $\left(1\right)$ is $m$ and that of line $\left(2\right)$ is $\frac{1}{3}$

Now,

$\frac{\frac{1}{3}-m}{1+m\times \frac{1}{3}}=\tan {45}^{\circ }$

$\Rightarrow -m+\frac{1}{3}=1+\frac{m}{3}$

$\Rightarrow m=-\frac{1}{2}$.

Then equation $\left(1\right)$ becomes $2y+x=c$.

This line passes through $(1,3)$

$\Rightarrow c=7$

Hence, the required equation is $2y+x=7$