Question

Lines $7x-2y+10=0$ and $7x+2y-10=0$ forms an isosceles triangle with the line $y=2$. Area of this triangle is equal to.

• A. $\frac{15}{7}$
• B. $\frac{10}{7}$
• C. $\frac{18}{7}$
• D. $\frac{8}{7}$

Answer 1

The correct option is C $\frac{18}{7}$

Line $L_1$ and line $L_2$ intersect at $y=5,x=0$

$L_1,x=\frac{2y}{7}-\frac{10}{7}$

and $L_2,x=\frac{10}{7}-\frac{2y}{7}$

hence area bounded by both line and $y=0$ is

$A={\int }_2^5(\frac{2y}{7}-\frac{10}{7})+{\int }_2^5(\frac{10}{7}-\frac{2y}{7})$

$A=2\times \left[{\int }_2^5\right(\frac{2y}{7}-\frac{10}{7}\left)\right]$

on solving and putting upper and lower limit we will get

$A=\frac{18}{7}$