Math, asked by pavansurwase125, 1 year ago

The point A(1-2) B(2,3) C(K,2) and d (-4,-3) ar the vortices of a parallogram find the value of k

Answers

Answered by sujnat505
7

Answer:

K=1

Step-by-step explanation:

Given: A(1-2) B(2,3) C(K,2) and d (-4,-3) are the vertices of a parallelogram

==>AB=CD

Now,

By distance formula,

AB=√[(X1-X2)²+(Y1-Y2)²]

=√[(1-2)² + (-2-3)²]

=√[1+25]

=√26


Now,

By distance formula,

cd=√[(X1-X2)²+(Y1-Y2)²]

=√[(K-(-4))² + (-2-(-3))²]

=√[(K+4)² + 1]


AB=CD

==>√26=√[(K+4)² + 1]

==>26=K²+16+8K+1

==>K²+8K-9=0

==>K² + 9K - 1K -9=0

==>K(K+9) -1(K+9) =0

==>K=1, siNCE K CANNOT BE NEGATIVE

Answered by kiranvkurienp90j3j
3

Answer:

Step-by-step explanation:

Given: A(1-2) B(2,3) C(K,2) and d (-4,-3) are the vertices of a parallelogram

==>AB=CD

Now,

By distance formula,

AB=√[(X1-X2)²+(Y1-Y2)²]

=√[(1-2)² + (-2-3)²]

=√[1+25]

=√26

Now,

By distance formula,

cd=√[(X1-X2)²+(Y1-Y2)²]

=√[(K-(-4))² + (-2-(-3))²]

=√[(K+4)² + 1]

AB=CD

==>√26=√[(K+4)² + 1]

==>26=K²+16+8K+1

==>K²+8K-9=0

==>K² + 9K - 1K -9=0

==>K(K+9) -1(K+9) =0

==>K=1, SINCE K CANNOT BE NEGATIVE

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