The points $A (1, - 2), B (2, 3), C (k, 2)$ and $D (- 4, - 3)$ are the vertices of a parallelogram. Find the value of $k$ .

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Solution

$\begin{matrix} A B = C D u sin g d i s tan c e f o r m u l a \sqrt{{(2 - 1)}^{2} + {(3 + 2)}^{2}} = \sqrt{{(k + 4)}^{2} + {(3 - 2)}^{2}} \sqrt{26} = \sqrt{k^{2} + 8 k + 41} s q u a r i n g b o t h s i d e 26 = k^{2} + 8 k + 41 k^{2} + 8 k + 15 = 0 k (k + 5) + 3 (k + 5) = 0 k = (- 3, - 5) \end{matrix}$

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The points A(1,−2),B(2,3),C(k,2) and D(−4,−3) are the vertices of a parallelogram. Find the value of k.

AB=CDusingdistanceformula√(2−1)2+(3+2)2=√(k+4)2+(3−2)2√26=√k2+8k+41squaringbothside26=k2+8k+41k2+8k+15=0k(k+5)+3(k+5)=0k=(−3,−5)

The points $A (1, - 2), B (2, 3), C (k, 2)$ and $D (- 4, - 3)$ are the vertices of a parallelogram. Find the value of $k$ .