Question

The vertices A(x, 2); B (3,4) and C (3, y), D (4,2) forms a parallelogram then x +y is
​

Answer 1

Given : The vertices A(x, 2); B (3,4) and C (3, y), D (4,2) forms a parallelogram

To Find : x + y

Solution:

A(x, 2); B (3,4) and C (3, y), D (4,2)

Opposite sides of a parallelogram are parallel

Parallel  lines have same slope

Slope of AB = Slope of CD

=> ( 2 - 4)/(x - 3)  =   (y  - 2)/(3 - 4)

=> -2/(x - 3) = (y - 2)/-1

=> 2 = (x - 3)(y - 2)

Slope of AC = Slope of BD

(y - 2)/(3 - x) = (2 - 4)/(4 - 3)

=> (y - 2)/(3 - x) = -2

=> y - 2 = - 2(3 - x)

=> y - 2 = 2(x - 3)

2 = (x - 3)(y - 2)

=> 2 = = (x - 3) 2(x - 3)

=> (x - 3)² = 1

=> x - 3 = ± 1

=> x = 4  , 2

x = 4 not possible as then A & D will be same point

=> x = 2

=> y - 2 = 2(x - 3)

=> y  - 2 = 2(2 - 3)

=> y = 0

x + y = 2 + 0 = 2

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Answer 2

$\textbf{Given:}$

$\textsf{Vertices of parallelogram ABCD are}$

$\mathsf{A(x,2),B(3,4),C(3,y)\;and\;D(4,2)}$

$\textbf{To find:}$

$\textsf{The value of x+y}$

$\textbf{Solution:}$

$\textsf{We know that, an important property of parallelogram}$

$\boxed{\textsf{Diagonals of parallelogram bisect each other}}$

$\implies\textsf{Midpoint of diagonal AC=Midpoint of diagonal BD}$

$\implies\mathsf{\left(\dfrac{x+3}{2},\dfrac{2+y}{2}\right)=\left(\dfrac{3+4}{2},\dfrac{4+2}{2}\right)}$

$\implies\mathsf{\left(\dfrac{x+3}{2},\dfrac{2+y}{2}\right)=\left(\dfrac{7}{2},\dfrac{6}{2}\right)}$

$\implies\mathsf{\left(\dfrac{x+3}{2},\dfrac{2+y}{2}\right)=\left(\dfrac{7}{2},3\right)}$

$\textsf{Equating corresponding co-ordinates on bothsides, we get}$

$\mathsf{\dfrac{x+3}{2}=\dfrac{7}{2}\;\;and\;\;\dfrac{2+y}{2}=3}$

$\mathsf{x+3=7\;\;and\;\;2+y=6}$

$\mathsf{x=7-3\;\;and\;\;y=6-2}$

$\mathsf{x=4\;\;and\;\;y=4}$

$\implies\boxed{\mathsf{x+y=8}}$

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