Question

The point 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 AB.

Answer 1

We know that,

Diagonals a parallelogram

bisects each other

So, mid points of

AC = mid points of BD

$(\frac{k+1}{2},\frac{2-2}{2})=(\frac{-4+2}{2},\frac{-3+3}{2})$

$\frac{k+1}{2}=1$

$k+1=-2$

$k=-3$

Now area of $△ABC$

$\Rightarrow \frac{1}{2}\left(x_1\right(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2\left)\right)$

$\Rightarrow \frac{1}{2}\left(1\right(3-2)+2(2+2)+(-3\left)\right(-2-3\left)\right)$

$\Rightarrow \frac{1}{2}\left(1\right(1)+2(4)-3(-5\left)\right)$

$\Rightarrow \frac{1}{2}(1+8+15)=\frac{24}{2}=12uni{t}^2$

area (parallelogram ABCD) $=2\times area\left(△ABC\right)$

$=2\times 12$

$=24uni{t}^2$

Area of Parallelogram $Base\times weight$

height $=\frac{24}{Base}$

So $DC=\sqrt{(x_2+x_1{)}^2+(y_2-y_1{)}^2}$

$=\sqrt{(3+4{)}^2+(2+3{)}^2}=\sqrt{26}$ units

height $=\frac{24}{\sqrt{26}}$