Question

The three vertices of a parallelogram ABCD taken in order A (-1,0) B(3,1) C(2,2) D (x,y) find the height of parallelogram with AD as it's base

Answer 1

hope this answer helps you

Answer 2

Answer:

Concept:

A simple (non-self-intersecting) quadrilateral with two sets of parallel sides is known as a parallelogram in Euclidean geometry. A parallelogram's facing or opposing sides are of equal length, and its opposing angles are of similar size. The Euclidean parallel postulate or one of its equivalent formulations must be used in order to demonstrate the congruence of opposed sides and opposite angles because both conditions are a direct result of this postulate.

Given :

A = ( -1,0 )

B = ( 3 ,1 )

C = ( 2, 2 )

D = ( x, y)

To Find:

The Height of parallelogram with AD as it's base.

Solution :

Midpoint of AC = $[\frac{(-1 + 2)}{ 2}$  , $\frac{( 0 + 2)}{ 2} ]$

Midpoint of AC  = $[\frac{( 1)}{ 2} , 1 ]$

Midpoint of AC  =  Midpoint of BD

Midpoint of BD = $\left ( \frac{ 3 + x}{2} , \frac{1+ y}{2} \right )$

Midpoint of AC  =   Midpoint of BD

      $[\frac{( 1)}{ 2} , 1 ]$           =   $\left ( \frac{ 3 + x}{2} , \frac{1+ y}{2} \right )$

Finding value of $x$

 $\frac{1}{2}$  = $\frac{ 3 + x}{2}$

$x = -2$

Finding value of y

$\frac{1+ y}{2}$  = 1

$y = 2 - 1$

$y = 1$

∴ D = ( -2 , 1)

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