Question

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

• A. $\frac{15}{7}$ sq. units
• B. $\frac{10}{7}$ sq. units
• C. $\frac{18}{7}$ sq. units
• D. none of these

Answer 1

Answer: (C)

$\frac{18}{7}$ sq. units

$7x-2y+10=0$ ...(i)

$7x+2y-10=0$ ...(ii)

Solving the above two equations, gives us

$14x=0$

i.e. $x=0$

$\Rightarrow y=5$

Hence, one vertex is $(0,5)$

Since the third side is $y=2$.

The other two vertices are $(\frac{-6}{7},2)$ and $(\frac{6}{7},2)$

Hence the vertices of the triangle are

$A=(0,5)$

$B=(\frac{-6}{7},2)$

$C=(\frac{6}{7},2)$

Thus, the area by determinant method is

$=\frac{1}{2}|\frac{-12}{7}+\frac{30}{7}-(\frac{-30}{7}+\frac{12}{7})|$

$=\frac{1}{2}\left|\frac{36}{7}\right|$

$=\frac{18}{7}$ sq units.