Question

If the points A(1, -2), B( 2, 3). (a, 2) and D(-4, -3) forms a parallelogram, find the value
of 'a'​

Answer 1

Answer:

a = -3

Step-by-step explanation:

given, A(1, -2), B( 2, 3). C(a, 2) and D(-4, -3)  forms a parellogram

distance between P(x₁,x₂) and Q(y₁,y₂) ,d(PQ) is given by the formula,

d =$\sqrt{(x\subscx_{1} - x\subscx_{2} )^2 +(y\subscx_{1} - y\subscx_{2} )^2 }$

so,

d(AB) = $\sqrt{(1 - 2)^2 +(-2 - 3)^2 }$ = $\sqrt{(- 1)^2 +(-5)^2 }$ = $\sqrt{1 +25 }$ = $\sqrt{26 } units$

similarly,

d(BC) = $\sqrt{(2-a)^2 +(3-2)^2 }$ = $\sqrt{(2-a)^2 +(1)^2 }$ = $\sqrt{(2-a)^2 +1}$ units

d(CD) = $\sqrt{(a-( - 4))^2 +(2 - (-3))^2 }$ = $\sqrt{(a+4)^2 +(5)^2 }$ = $\sqrt{(a+4)^2 +25 } units$d(AD) = $\sqrt{(1-( - 4))^2 +(-2 - (-3))^2 }$ = $\sqrt{(1+4)^2 +(-2+3)^2 }$ = $\sqrt{25+1 }$ = $\sqrt{26 } units$

if AB and AD are equal, then all 4 sides are equal, (since opp sides are equal in a parallelogram )

side opp to AB = CD   , so  d(AB) = d(CD) => d(CD) = $\sqrt{26 } units$   --- (1)

side opp ot BC = AD   , so  d(BC) = d(AD) => d(BC) = $\sqrt{26 } units$   ---(2)

take, (1)

d(CD) = $\sqrt{26 } units$

=> $\sqrt{(a+4)^2 +25 }$ = $\sqrt{26 }$

=> (a+4)² + 25 = 26

=> (a+4)² = 26-25 = 1

=> a+4 = 1

=> a = 1-4 = -3

a = -3