If the points $A (1, - 2), B (2, 3), C (- 3, 2)$ and $D (- 1, - 3)$ are the vertices of $11 g$ ABCD with AB as base, find the height of the $11 g$ ABCD.

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Solution

$A (1, - 2)$ $B (2, 3)$ $C (- 3, 2)$ $D (- 1, - 3)$
We need height of the quadrilateral
$h_{1}$ perpendicular distance from $D$ to line $A B$ .
equation of line $A B$
$(y + 2) = \frac{(3 + 2)}{(2 - 1)} (x - 1)$
$(y + 2) = 5 (x - 1)$
$y = 5 x - 7$
perpendicular distance from $D$ to $Y = 5 x - 7$
$\Rightarrow \frac{- 3 - 5 (- 1) + 7}{\sqrt{1^{2} + 5^{2}}} \Rightarrow \frac{9}{\sqrt{26}} = h_{1}$
$h_{2} =$ perpendicular from $C$ to $Y = 5 x - 7$
$\Rightarrow \frac{2 + 7 - 5 (- 3)}{\sqrt{1^{2} + 5^{2}}} \Rightarrow \frac{29}{\sqrt{26}} = h_{2}$
$∴$ of $A B C D$ is $\frac{24}{\sqrt{26}}$

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If the points A(1,−2),B(2,3),C(−3,2) and D(−1,−3) are the vertices of 11g ABCD with AB as base, find the height of the 11g ABCD.

If the points $A (1, - 2), B (2, 3), C (- 3, 2)$ and $D (- 1, - 3)$ are the vertices of $11 g$ ABCD with AB as base, find the height of the $11 g$ ABCD.