Question

28.Threee vertices of a parallelogram ABCD are ( 1, 2 ) , ( 4, 3 ) ,( 6 , 6 ) . Find the 4th vertex.​

Answer 1

Answer:

3,5

Step-by-step explanation:

Use two points distance formula (as opp. sides are equal in parallelogram

And two point slope formula ( as opp. sides of a parallelogram have equal slopes

Answer 2

$\sf \huge \underline \blue{†Given†}$

Three vertices of a parallelogram are A( 1, 2 ) , B ( 4, 3 ) ,C( 6 , 6 )

$\sf \huge \underline \blue{†To \: \: Find†}$

4th vertex of parallelogram

$\sf \huge \underline \blue{†Concept†}$

In a parallelogram, diagonals bisect each other

$\sf \huge \underline \blue{†Formula\:\: Used†}$

$\bf \large \green{»P(x,y) = \left( \dfrac{x_1 + x_2}{2} \: \: \: \dfrac{y_1 + y_2}{2} \right)}$

$\sf \huge \underline \blue{†Solution†}$

O is the mid point of AC and BD

NOW,

In line AC

$\tt \leadsto \pink{P(x,y)= \left( \frac{1 + 6}{2} \: \: \: \frac{2 + 6}{2} \right )}$

$\tt \leadsto \pink{P(x,y)= \left( \frac{7}{2} \: \: \: \: \: \: ,\: \: \: \frac{8}{2} \right )}$

$\tt \leadsto \pink{P(x,y)= \left( \frac{7}{2} \: \: \: \: \: \: ,\: \: \: 4 \right )}$

In line BD

$\tt \leadsto \pink{ \left( \frac{4 + x}{2} \: \: \: \: \: \: ,\: \: \: \frac{3 + y}{2} \right ) = \left( \frac{7}{2} \: \: \: \: \: \: ,\: \: \: \frac{8}{2} \right )}$

$\tt \leadsto \pink{ \left( \frac{4 + x}{2} \right) = \left( \frac{7}{2} \right)\: \: \: \: \: \: ,\: \: \: \left (\frac{3 + y}{2} \right ) = \: \: \: \: \: \: \: \: \: \left( \frac{8}{2} \right )}$

$\tt \leadsto \pink{4 + x = 7 \: \: and \: \: 3 + y = 8}$

$\tt \leadsto \pink{x = 3 \: \: and \: \: y = 5}$

•°• The fourth vertex of parallelogram is D (3,5)