Answer to Question #228765 in Analytic Geometry for Bless

Question

A triangle is formed by the lines y=x, y=2x-1 and 3y-4x=1. Find the area of the triangle

Answer 1

Given equation are $y=x,y=2x-1,3y-4x=1$

Let $y=x...(1)$

$y=2x-1…(2)$

$3y-4x=1…(3)$

Solving (1) and (2), we get:

$(x,y)=(1,1)$

Solving (2) and (3), we get:

$3(2x-1)-4x=1$

$⇒2x=4$ 

$⇒x=2$

$∴y=3$

$(x,y)=(2,3)$

Solving (1) and (3), we get:

$(x,y)=(-1,-1)$

So, let coordinates of points A, B, C of $\triangle ABC$ be $(1,1),(2,3),(-1,-1)$ respectively.

We know that area of triangle with vertices $(x_1,y_1),(x_2,y_2),(x_3,y_3)$ is given as:

$\triangle =|\frac{1}{2} \begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix}|$

$\therefore$ Area of $\triangle ABC=|\frac{1}{2} \begin{vmatrix} 1 & 1 & 1 \\ 2 & 3 & 1 \\ -1 & -1 & 1 \end{vmatrix}|$

$=\frac{1}{2} [1(3+1)-1(2+1)+1(-2+3)]$

$=\frac{1}{2} [4-2+1]$

$=\frac{3}{2}$ square units.