Question

♦EASY♦

If (1, 2), (4, y), (x, 6) and (3, 5) are the vertices of a parallelogram taken in order, find x and y.

μ. DONT SPAM ¶​

Answer 1

AnswEr:

Let us Consider that A, B, C & D be the Points of the Parallelogram.

A(1,2), B(4,y), C(x,6) & D(3,5)

We know that, the diagonals of Parallelogram bisect each other.

Finding the value of x & y :-

$\dag\:\:\small\bold{\underline{\sf{\red{By\: Using\:Mid \:-\: Point\: Formula}}}}$

$:\implies\small{\underline{\boxed{\sf{\pink{\dfrac{x_1 \:+\; x_2}{2} \:,\: \dfrac{y_2\:+\;y_2}{2}}}}}}$

Diagonal AC

Here,

• x1 = 1
• x2 = x
• y1 = 2
• y2 = 6

★ Putting Values:-

$:\implies\sf\: \dfrac{1 \:+\:x}{2}\:,\:\dfrac{2\:+\:6}{2}$

$:\implies\sf \:\dfrac{1\:+\:x}{2}\:,\:4$

Now, Diagonal BC

Here,

• x1 = 4
• x2 = 3
• y1 = 5
• y2 = y

$:\implies\sf\:\dfrac{4 \:+\:3}{2}\:,\:\dfrac{5\:+\:y}{2}$

$:\implies\sf\: \dfrac{7}{2}\: , \: \dfrac{5\:+\:y}{2}$

Now, Mid points of Both Diagonals

$:\implies\sf\:\dfrac{1\:+\:X}{2} \:=\: \dfrac{7}{2} \: and \; 4 = \dfrac{5 \:+\: y}{2}$

$:\implies\sf\: x + 1 = 7$

$:\implies\sf\: x = 7 - 1$

$:\implies\large{\underline{\boxed{\sf{\blue{x\:=\:6}}}}}$

$\rule{150}2$

$:\implies\sf\: 5 + y = 8$

$:\implies\sf\: y = 8 - 5$

$:\implies\large{\underline{\boxed{\sf{\blue{y\:=\:3}}}}}$

$\bold{\underline{\sf{\pink{Hence,\: Value\: of \: x \: and \: y \: is \: 6 \: and \: 3.}}}}$

Answer 2

Answer:

Let:-

• P (1,2)
• Q (4,y)
• R (x,6)
• S (3,5)

Concept:- The diagonals of a parallelogram bisect eachother.

Hence, the midpoint of diagonal PR and the midpoint of diagonal QS will be the pint of intersection of both the diagonals.

Formula used:-

$\mapsto\tt{\green{\dfrac{x_1 + x_2}{2},\dfrac{y_1 + y_2}{2}}}$

Now, finding the midpoint of PR.

Here,

• x1 = 1
• y1 = 2
• x2 = x
• y2 = 6

Putting the values in the formula:-

$\mapsto\tt{\dfrac{1 + x}{2},\dfrac{2 + 6}{2}}$

$\mapsto\tt{\dfrac{1 + x}{2},\dfrac{8}{2}}$

$\mapsto\tt{\dfrac{1 + x}{2},4}$

Now, finding the midpoint of QS.

Here,

• Here,x1 = 4
• y1 = y
• x2 = 3
• y2 = 5

Putting the values in the formula:-

$\mapsto\tt{\dfrac{4 + 3}{2},\dfrac{y + 5}{2}}$

$\mapsto\tt{\dfrac{7}{2},\dfrac{y + 5}{2}}$

Now, as the midpoint of both the diagonals are same.

Hence,

$\mapsto\tt{\dfrac{7}{2}=\dfrac{1+x}{2},\dfrac{y + 5}{2}=4}$

$\mapsto\tt{14=2(1+x),y + 5=8}$

$\mapsto\tt{7=1+x,y =8-5}$

$\mapsto\tt{x=7-1,y =3}$

$\huge\mapsto\tt{\green{x=6,y =3}}$

$\rule{200}4$