Question

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

• A. Centroid
• B. Incentre
• C. Orthocentre
• D. None of these

Answer 1

Answer: (C)

Orthocentre

Given

$L_1;4x-7y+10=0$; slope($m_1=4/7$)

$L_2;7x+4y=15$; slope($m_2=-7/4$)

$L_3;x+y=5$; slope($m_3=-1$)

Product of slope of $L_1$ and $L_2$ is $-1$, so they are perpendicular to each other.

So triangle form by these three lines are right angled triangle.

Also in right angled triangle orthocentre lies on the vertex which is right angle.

Hence point of intersection of $L_1$ and $L_2$ gives the orthocentre.

So solving equations of line(1) and line(2), we get its intersection as $H(1,2)$