Question

find the values of a and b are vertices of a parallelogram (-8,10), (2,8), (a,3) and (4,b) taken in order.
With full explanation
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Answer 1

$\huge{ \blue{\underline{ \overline{ \boxed{ \orange{ \mathbb{ANSWER}}}}}}}$

Let ABCD be a parellogram whose vertices (-8,10), (2,8), (a,3) and (4,b) are taken in order.

TO FIND :

Values of 'a' and 'b'

SOLUTION :

Let A = (-8,10), B = (2,8), C = (a,3) and D = (4,b).

Let A = (-8,10), B = (2,8), C = (a,3) and D = (4,b).Let D(x,y) be the point of intersection of the diagonals of the parellogram.

Now, we know that the diagonals of parellogram bisect each other.

Now, from diagonal AC, D(x,y) is the mid point of AC.

$\therefore \: \sf \: D(x,y) = ( \frac{ x_1 \: + \: x_3}{2} , \: \frac{ y_1 \: + \: y_3}{2} )$

$\sf \: D(x,y) = ( \frac{ - 8 \: + \: a}{2} , \: \frac{10 \: + \: 3}{2} ) \: \: \: \: \: \: \rightarrow \: (1)$

Also, from diagonal BD, D(x,y) is the mid point of BD.

$\therefore \: \sf \: D(x,y) = ( \frac{ x_2 \: + \: x_4}{2} , \: \frac{ y_2 \: + \: y_4}{2} )$

$\sf \: D(x,y) = ( \frac{ 4 \: + \: 2}{2} , \: \frac{b \: + \: 8}{2} ) \: \: \: \: \: \: \rightarrow \: (2)$

Now, from (1) and (2) we get,

$\implies \: ( \frac{ - 8 \: + \: a}{2} , \: \frac{10 \: + \: 3}{2} ) = \: ( \frac{ 4 \: + \: 2}{2} , \: \frac{b \: + \: 8}{2} )$

On comparing we get,

$\implies \: \sf{ \frac{ - 8 \: + \: a}{2} \: = \frac{ 4 \: + \: 2}{2} \: \: \: and \: \: \: \: \frac{b \: + \: 8}{2} \: = \: \frac{10 \: + \: 3}{2} }$

$\implies \: \sf{ { - 8 + a} \: = { 4 + 2}\: \: \: \: and \: \: \: \: \: {b + 8} = {10 + 3} }$

$\implies \: \sf{ a \: = 14\: \: \: \: and \: \: \: \: \: b = 5 }$

$\boxed{ \: \sf \green{ Required \: numbers \: are : \: a \: = 14 \: \: \: and \: \: \: b = 5 \: }}$

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