Question

The area of a triangle formed by the lines $x+y-3=0,x-3y+9=0$ and $3x-2y+1=0$ is -

• A. $\frac{20}{7}$ sq. units
• B. None of these
• C. $\frac{5}{7}$ sq. units
• D. $\frac{10}{7}$ sq. units

Answer 1

The correct option is D $\frac{10}{7}$ sq. units

Solving equation we can write,
$x+y-3=0\left(1\right)$
$x-3y+9=0-\left(2\right)$
$3x-2y+1=0(-3)$



Since these $3$ lines form a triangle, the point where $\left(1\right)$ and $\left(2\right)$ intersect is called the vertex A of the triangle. The point where $(2$) and $\left(3\right)$ intersect is called the vertex B and the point where $\left(1\right)$ and $\left(3\right)$ intersect is called the vertex C of the tnangle.
Subtracting equation $\left(2\right)$ from equation $\left(1\right)$ we get, $y=3$
Substituting the value of $y=3$ in $\left(1\right)$ we get $x=0$
$\therefore$ Coordinates of Vertex $A=\left(0.3\right)$
Similarly multiplying equation $\left(1\right)$ with $3$ and subtracting $\left(3\right)$ from $\left(1\right)$ we get $y=2$
Substituting the value of $y=2$ in $\left(3\right)$ we get $x=1$
$\therefore$ Coordinates of Vertex C = $(1,2)$
Similarly multiplying equation $\left(3\right)$ with $3$ and substracting $\left(3\right)$ from $\left(2\right)$ we get $x=\frac{15}{7}$

Substituting the value of $x$ in $\left(2\right)$ we get $y=\frac{26}{7}$

$\therefore$ Coordinates of Vertex $B=(\frac{15}{7},\frac{26}{7})$

So, we have the coordinates $A,B,C$. Now we have to find the area using the direct formula. Putting the values in the direct formula

$Area=\frac{1}{2}\left[x_1\right(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2\left)\right]$
Substituting the values, we get

$\Rightarrow Area=\frac{1}{2}\left[0\right(2-\frac{26}{7})+1(\frac{26}{7}-3)+\frac{15}{7}(3-2\left)\right]$

$\Rightarrow Area=\frac{1}{2}[0+\frac{5}{7}+\frac{15}{7}]$

$\Rightarrow Area=\frac{10}{7}sq.unit$

Hence, the Area of the triagle is $\frac{10}{7}$ sq. units