Question

The three lines $4x-7y+10=0,x+y=5$ and $7x+4y=15$ form the sides of a triangle. Then the point $(1,2)$ is its

• A. centroid
• B. incentre
• C. orthocentre
• D. none of these

Answer 1

The correct option is C orthocentre

Given lines

$4x-7y+10=0----\left(1\right)$

$x+y-5=0----\left(2\right)$

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

On solving eq (1) and (3)

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

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

$65x=65\Rightarrow x=1$

From eq (1)

$4-7y=-10$

$-7y=-14$

$y=2$

Point of intersection is $(1,2)$

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

Hence the orthocentre of Right angle triangle is the point at which ${90}^0$ angle is formed

Hence $(1,2)$ is orthocentre