Question

One side of a rectangle lies along the line $4x+7y+5=0$ and two vertices are $(-3,1),(1,1)$ then the remaining vertices are,

• A. $\left(\frac{1}{65}\frac{-47}{65}\right),\left(\frac{-131}{65},\frac{177}{65}\right)$
• B. $\left(\frac{-1}{65},\frac{47}{65}\right),(\frac{-131}{65},\frac{177}{65})$
• C. $\left(\frac{1}{65},\frac{-47}{65}\right),(\frac{131}{65},\frac{-177}{65})$
• D. $(1,-47),(131,47)$

Answer 1

Answer: (C)

$\left(\frac{1}{65},\frac{-47}{65}\right),(\frac{131}{65},\frac{-177}{65})$

Since, the side $AB$ is perpendicular to $AD$

$\therefore$ its equation is of the form $7x-4y+\lambda =0$

Since, it passes through $(-3,1)$

$\therefore 7(-3)-4\left(1\right)+\lambda =0\Rightarrow \lambda =25$

$\therefore$ Equation of $AB$ is $7x-4y+25=0$

Now, $BC$ is parallel to $AD.$

Therefore, its equation is $4x+7y+\lambda =0$

Since, it passes through $(1,1).$

$\therefore 4\left(1\right)+7\left(1\right)+\lambda =0\to \lambda =-11$

$\therefore$ Equation of $BC$ is $4x+7y-11=0$

Now, equation of $DC$ is $7x-4y+\lambda =0$

$\Rightarrow 7\left(1\right)-4\left(1\right)+\lambda =0\Rightarrow \lambda =-3$

$\therefore 7x-4y-3=0$

Hence coordinate of remaining vertices is $(\frac{1}{65},\frac{-47}{65}),(\frac{131}{65},\frac{-177}{65})$