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 $k$ and the altitude of the parallelogram corresponding to the base $A B$ .

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Solution

$A B | | D C$

$∴ \frac{- 2 - 3}{1 - 2} = \frac{- 3 - 2}{- 4 - k}$

$5 (4 + K) = 5$

$4 + k = 1$

$k = - 3$

Slope of $A B = \frac{- 2 - 3}{1 - 2} = 5$

Slope of attitude $= \frac{- 1}{5}$

attitude $= \frac{- 1}{5} (x - (- 3)) = (y - 2)$

passing through $= c$

$\Rightarrow 5 y - 10 = - x - 3$

$5 y + x = 7$

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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 and the altitude of the parallelogram corresponding to the base AB.

AB||DC∴−2−31−2=−3−2−4−k5(4+K)=54+k=1k=−3Slope of AB=−2−31−2=5Slope of attitude =−15attitude =−15(x−(−3))=(y−2)passing through =c⇒5y−10=−x−35y+x=7

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 $k$ and the altitude of the parallelogram corresponding to the base $A B$ .