Question

The orthocenter of the triangle whose sides are given by $4x-7y+10=0,x+y-5=0$ and $7x+4y-15=0$ is

• A. $(-1,-2)$
• B. $(1,-2)$
• C. $(-1,2)$
• D. $(1,2)$

Answer 1

The correct option is D $(1,2)$

Given lines

$4x-7y+10=0$

$\Rightarrow 7y=4x+10$

$\Rightarrow y=\frac{4x+10}{7}$ .....$\left(1\right)$

$x+y-5=0$

$\Rightarrow y=5-x$ .....$\left(2\right)$

$7x+4y-15=0$ .....$\left(3\right)$

Substituting $\left(1\right)$ in $\left(3\right)$ we get

$7x+4y-15=0$

$7x+4\left(\frac{4x+10}{7}\right)-15=0$

$\Rightarrow 49x+16x+40-105=0$

$\Rightarrow 65x=65\Rightarrow x=1$

From eqn$\left(1\right)$

$4-7y=-10$

$\Rightarrow -7y=-14$

$\therefore y=2$

Point of intersection is $(1,2)$

Here Equation $\left(1\right)$ and $\left(3\right)$ are perpendicular So the lines formed right angle triangle

Hence the orthocentre of Right angle triangle is the point at which ${90}^{\circ }$ angle is formed

Hence $(1,2)$ is orthocentre