Question

The area (in sq. units) of the triangle formed by the lines 2x + 3y = 3, x + y = 3 and y + 1 = 0 is

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Answer 1

The area of the triangle formed by the lines 2x + 3y = 3, x + y = 3 and y + 1 = 0 is 1 sq units

Step-by-step explanation:

The given lines are

$L_1: 2x+3y=3$  ........... (1)

$L_2: x+y=3$      ............. (2)

$L_3: y+1=0$     ............... (3)

First we need to find the point of intersection of these lines

Solving the simultaneous equation of lines $L_1$ and $L_2$

We get point of intersection of lines $L_1$ and $L_2$ as $(6,-3)$

Solving the simultaneous equation of lines $L_2$ and $L_3$

We get point of intersection of lines $L_2$ and $L_3$ as $(4,-1)$

Solving the simultaneous equation of lines $L_1$ and $L_3$

We get point of intersection of lines $L_1$ and $L_3$ as $(3,-1)$

We know that area of a triangle formed by three points $(x_1,y_1)$, $(x_2,y_2)$ and $(x_3,y_3)$ is given by

$\boxed{A=\frac{1}{2}|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)|}$

Therefore, the area of triangle formed by points $(6,-3)$, $(4,-1)$ and $(3, -1)$ is given by

$A=\frac{1}{2}|6[-1-(-1)]+4[-1-(-3)]+3[-3-(-1)]|$

$\implies A=\frac{1}{2}|0+8-6|$

$\implies A=\frac{1}{2}\times 2$

$\implies A=1$ sq units

Hope this answer is helpful.

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