Question

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

• A. $\frac{16}{7}$ sq. units
• B. $\frac{10}{7}$ sq. units
• C. $4$ sq. units
• D. $9$ sq. units

Answer 1

Answer: (B)

$\frac{10}{7}$ sq. units

Equation of sides is given by

$x+y=3$-------(i)

$3x-2y+1=0$---(ii)

$x-3y+9=0$-----(iii)

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

$=\frac{1}{2}|1(3-\frac{26}{7})+0(y_3-y_1)+\frac{15}{7}(2-3\left)\right|$

$=\frac{1}{2}|-\frac{5}{7}-\frac{15}{7}|$

$=\frac{1}{2}\times \frac{20}{7}=\frac{10}{7}$

$\therefore △=\frac{10}{7}squnit$