Question

If the points A(1,-2) , B(2,3) , C(-3,2) and D(-4,-3) are the vertices of a parallelogram ABCD, then taking AB as the base, find the height of the parallelogram.

Answer 1

Since llgm ABCD and ∆ABC lie on the same base and the same parallels, thus,
​area(||gmABCD)=2area(∆ABC)
By using the area formula we have,

Area of a triangle=½|x1(y​2-y3)+x2(y​3-y1)+x3(y1​-y2)
=½|1(3-2)+2(2+2)+(-3)(-2-3)|
=½|1+8+15|
=½×24=12 square units
Therefore, area(||gm ABCD)= 2×12 =24 square units
Now,
Using the distance formula we have,
AB=√(x2-x1)2+(y2-y1)2
=√(2-1)2+(3+2)2
=√1+25
=√26 units
We know that,
Area of a parallelogram=height×base
=>area(llgm ABCD)=height ×AB
=>24=height ×√26
=>height=24/√26