Question

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

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

Answer 1

Answer: (A)

$\frac{16}{7}$ sq. unit

To find $B$

$3x-2y=-1$ and $x+y=3$

solving above two equations we get

$y=2$ and $x=1$

$B=(1,2)$

To find $C$

$3x-2y=-1$ and $x-3y=-9$

solving above two equations we get

$y=27/6$ and $x=15/7$

$C=(\frac{15}{7},\frac{27}{6})$

$s=\frac{perimeteroftriangle}{2}$

$a=AB=\sqrt{{(1-0)}^2+{(2-3)}^2}=\sqrt{1+1}=\sqrt{2}b=BC=\sqrt{{(\frac{15}{7}-1)}^2+{(\frac{26}{7}-2)}^2}=\sqrt{{\left(\frac{8}{7}\right)}^2+{\left(\frac{12}{7}\right)}^2}=\frac{4\sqrt{13}}{7}c=CA=\sqrt{{(\frac{15}{7}-0)}^2+{(\frac{26}{7}-3)}^2}=\sqrt{{\left(\frac{15}{7}\right)}^2+{\left(\frac{5}{7}\right)}^2}=\frac{5\sqrt{10}}{7}s=\frac{a+b+c}{2}s=\frac{\sqrt{2}+\frac{4\sqrt{13}}{7}+\frac{5\sqrt{10}}{7}}{2}s=\frac{14\sqrt{2}+4\sqrt{13}+5\sqrt{10}}{14}s-a=\frac{-7\sqrt{2}+4\sqrt{13}+5\sqrt{10}}{14}s-b=\frac{14\sqrt{2}-4\sqrt{13}+5\sqrt{10}}{14}s-c=\frac{14\sqrt{2}+4\sqrt{13}-5\sqrt{10}}{14}A=\sqrt{s(s-a)(s-b)(s-c)}A=\sqrt{\left(\frac{-7\sqrt{2}+4\sqrt{13}+5\sqrt{10}}{14}\right)\left(\frac{14\sqrt{2}-4\sqrt{13}+5\sqrt{10}}{14}\right)\left(\frac{14\sqrt{2}+4\sqrt{13}-5\sqrt{10}}{14}\right)}$

Multiplying and solving

$A=\frac{16}{7}squnit$