If three vertices of a square are $(-4, 3)$,$(10, 5)$,and $(12, -9)$,then find the fourth vertex of the square.

  • A
    $(-21, -10)$
  • B
    $(-2, -11)$
  • C
    $(-5, -9)$
  • D
    $(-3, -1)$

Explore More

Similar Questions

The area of a triangle having vertices $A(0,0), B(4,0)$ and $C(0,6)$ is:

$ABCD$ is a parallelogram with vertices $A(x_{1}, y_{1})$,$B(x_{2}, y_{2})$,and $C(x_{3}, y_{3})$. Find the coordinates of the fourth vertex $D$ in terms of $x_{1}, x_{2}, x_{3}, y_{1}, y_{2}$,and $y_{3}$.

Prove that the centroid of the triangle with vertices $(1, a), (2, b), (c^2, -3)$ does not lie on the $Y$-axis.

The distance between the points $(0, 5)$ and $(-5, 0)$ is

If $P (9a-2, -b)$ divides the line segment joining $A (3a+1, -3)$ and $B (8a, 5)$ in the ratio $3:1$,find the values of $a$ and $b$.

Difficult
View Solution

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