Question

The equations to the sides of a triangle are $x-3y=0,4x+3y=5$ and $3x+y=0$. The line $3x-4y=0$ passes through

• A. the centre
• B. the centroid
• C. the circumcentre
• D. the orthocentre of the triangle

Answer 1

Answer: (D)

the orthocentre of the triangle

Slope of line $4x+3y=5$ is $\frac{-4}{3}$

and Slope of line $3x-4y=0$ is $\frac{3}{4}$

Product of their Slopes is $\frac{3}{4}\ast \frac{-4}{3}=-1$

Hence, The line $3x-4y=0$ is Perpendicular to side $4x+3y=5$

And Orthocentre is the intersection of Perpendiculars

So, the line $3x-4y=0$ must have passed through the Orthocentre of the triangle