If $A(4,7)$,$B(-7,8)$,and $C(1,2)$ are the vertices of $\triangle ABC$,then the equation of the perpendicular bisector of the side $AB$ is

  • A
    $x-11y-24=0$
  • B
    $11x+y+24=0$
  • C
    $11x-y+24=0$
  • D
    $11x+y-24=0$

Explore More

Similar Questions

The point of intersection of the diagonals of the rectangle whose sides are contained in the lines $x=8, x=10, y=11$ and $y=12$ is

Let the circumcentre of a triangle with vertices $A(a, 3)$,$B(b, 5)$,and $C(a, b)$,where $ab > 0$,be $P(1, 1)$. If the line $AP$ intersects the line $BC$ at the point $Q(k_{1}, k_{2})$,then $k_{1} + k_{2}$ is equal to.

Area of the rhombus bounded by the four lines,$ax \pm by \pm c = 0$ is :

Difficult
View Solution

Two sides of a triangle have lengths $a \text{ cm}$ and $b \text{ cm}$. If the triangle is to have the maximum area,then the length of the median from the vertex containing the sides $a$ and $b$ is

For all $\alpha, \beta \in R$ and $\alpha \beta > 0$,the line $\alpha x + \beta y + \sqrt{\alpha \beta} = 0$ is such that it

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo