Without using the distance formula,show that | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(N/A) Let the points $(-2,-1), (4,0), (3,3),$ and $(-3,2)$ be denoted by $A, B, C,$ and $D$ respectively.The slope of $AB = \frac{0 - (-1)}{4 - (-2)}

Vedclass · Accepted Answer

Vedclass · Answer

(N/A) Let the points $(-2,-1), (4,0), (3,3),$ and $(-3,2)$ be denoted by $A, B, C,$ and $D$ respectively.
The slope of $AB = \frac{0 - (-1)}{4 - (-2)} = \frac{1}{6}$.
The slope of $CD = \frac{2 - 3}{-3 - 3} = \frac{-1}{-6} = \frac{1}{6}$.
Since the slope of $AB = \text{slope of } CD$,$AB$ is parallel to $CD$.
Now,the slope of $BC = \frac{3 - 0}{3 - 4} = \frac{3}{-1} = -3$.
The slope of $AD = \frac{2 - (-1)}{-3 - (-2)} = \frac{3}{-1} = -3$.
Since the slope of $BC = \text{slope of } AD$,$BC$ is parallel to $AD$.
Since both pairs of opposite sides are parallel,the quadrilateral $ABCD$ is a parallelogram.

Without using the distance formula,show that the points $(-2,-1), (4,0), (3,3),$ and $(-3,2)$ are the vertices of a parallelogram.

Similar Questions

The orthocentre of the triangle formed by the lines $4x - 7y + 10 = 0$,$x + y = 5$,and $7x + 4y = 15$ is

If a straight line perpendicular to $2x - 3y + 7 = 0$ forms a triangle with the coordinate axes whose area is $3 \text{ sq. units}$,then the equation of the straight line is:

The triangle formed by joining the points $P(2, 7)$,$Q(4, -1)$,and $R(-2, 6)$ is:

If in a parallelogram $ABDC$,the coordinates of $A, B$ and $C$ are respectively $(1, 2), (3, 4)$ and $(2, 5)$,then the equation of the diagonal $AD$ is

The coordinates of three points $A(-4, 0)$,$B(2, 1)$,and $C(3, 1)$ determine the vertices of an isosceles trapezium $ABCD$. The coordinates of the vertex $D$ are:

Vedclass Test Series

Exam Paper Generator

Online Exam Module