$A(0,2)$ and $C(6,4)$ are opposite vertices of square $ABCD$. The sum of the slopes of the sides passing through vertex $A$ is:

  • A
    $-\frac{1}{2}$
  • B
    $1.5$
  • C
    $2$
  • D
    $-2$

Explore More

Similar Questions

All points lying inside the triangle formed by the points $(1, 3)$,$(5, 0)$,and $(-1, 2)$ satisfy which of the following inequalities?

The circumcentre of the triangle formed by the lines $xy+2x+2y+4=0$ and $x+y+2=0$ is

The area enclosed by the graphs of $|x + y| = 2$ and $|x| = 1$ is

The equations of the sides of a triangle are $x - 2y = 0$,$4x + 3y = 5$,and $2x + y = 0$. Through which point does the line $3y - 4x = 0$ pass?

Difficult
View Solution

Let $\alpha, \beta, \gamma, \delta \in \mathbb{Z}$ and let $A(\alpha, \beta), B(1, 0), C(\gamma, \delta)$ and $D(1, 2)$ be the vertices of a parallelogram $ABCD$. If $AB = \sqrt{10}$ and the points $A$ and $C$ lie on the line $3y = 2x + 1$,then $2(\alpha + \beta + \gamma + \delta)$ is equal to

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