Question

A(6, 1),B(8, 2) and C(9,4) are the vertices of a parallelogram ABCD. If E is the midpoint of BD, find the area of $△ABE$.

Answer 1

Given, ACBD is a parallelogram.AC and BD are the diagonal intersecting in E.

Let the coordinates of D be (x, y).

We know that, diagonals of parallelogram bisect each other.

∴ Mid-point of BD = Mid point of AC

$(\frac{x+8}{2},\frac{y+2}{2})=(\frac{9+6}{2},\frac{1+4}{2})(\frac{x+8}{2},\frac{y+2}{2})=(\frac{15}{2},\frac{5}{2})\frac{x+8}{2}=\frac{15}{2}and\frac{y+2}{2}=\frac{5}{2}2x+16=30and2y+4=102x=30-16=14and2y=10-4=6x=7andy=3$

Coordinates of D is (7,3)

E is the midpoint of BD,

i.e, $E(p,q)=(\frac{8+7}{2},\frac{3+2}{2})=(\frac{15}{2},\frac{5}{2})$

We know area of a triangle with vertices $(x_1,y_1),(x_2,y_2),(x_3,y_3)$ is

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

Area of $△ABE$ with vertices $A(6,1),B(8,2),C(\frac{15}{2},\frac{5}{2})$ is,

$Area=\frac{1}{2}\left[x_1\right(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2\left)\right]=\frac{1}{2}\left[6\right(2-\frac{5}{2})+8(\frac{5}{2}-1)+\frac{15}{2}(1-2\left)\right]=\frac{1}{2}[6\times \frac{-1}{2}+8\times \frac{3}{2}+\frac{15}{2}\times -1]=\frac{1}{2}[-3+12-\frac{15}{2}]=\frac{1}{2}[9-\frac{15}{2}]=\frac{1}{2}\left[\frac{18-15}{2}\right]=\frac{1}{2}\times \frac{3}{2}=\frac{3}{4}$

Area of $△ABE=\frac{3}{4}sq.units$