The area of triangle formed by the lines x + | vedclass

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Question

To find $B$

$3 x-2 y=-1$ and $x+y=3$

solving above two equations we get $y=2$ and $x=1$

$B=(1,2)$

To find $C$

$3 x-2 y=-1$ and $x-3 y=-

Vedclass · Accepted Answer

$\frac{{16}}{7}\,$ sq. units

Vedclass · Answer

To find $B$

$3 x-2 y=-1$ and $x+y=3$

solving above two equations we get $y=2$ and $x=1$

$B=(1,2)$

To find $C$

$3 x-2 y=-1$ and $x-3 y=-9$

solving above two equations we get $y=27 / 6$ and $x=15 / 7$

$C=\left(\frac{15}{7}, \frac{27}{6}ight)$

$s=\frac{	ext {perimeteroftriangle}}{2}$

$a=A B=\sqrt{(1-0)^{2}+(2-3)^{2}}$

$=\sqrt{1+1}$

$=\sqrt{2}$

$b=B C=\sqrt{\left(\frac{15}{7}-1ight)^{2}+\left(\frac{26}{7}-2ight)^{2}}$

$=\sqrt{\left(\frac{8}{7}ight)^{2}+\left(\frac{12}{7}ight)^{2}}$

$\frac{4 \sqrt{13}}{7}$

$c=C A=\sqrt{\left(\frac{15}{7}-0ight)^{2}+\left(\frac{26}{7}-3ight)^{2}}$

$=\sqrt{\left(\frac{15}{7}ight)^{2}+\left(\frac{5}{7}ight)^{2}}$

$=\frac{5 \sqrt{10}}{7}$

$s=\frac{a+b+c}{2}$

$s=\frac{\sqrt{2}+\frac{4 \sqrt{13}}{7}+\frac{5 \sqrt{10}}{7}}{2}$

$s=\frac{14 \sqrt{2}+4 \sqrt{13}+5 \sqrt{10}}{14}$

$s-a=\frac{-7 \sqrt{2}+4 \sqrt{13}+5 \sqrt{10}}{14}$

$s-b=\frac{14 \sqrt{2}-4 \sqrt{13}+5 \sqrt{10}}{14}$

$s-c=\frac{14 \sqrt{2}+4 \sqrt{13}-5 \sqrt{10}}{14}$

$A=\sqrt{s(s-a)(s-b)(s-c)}$

$A=\sqrt{\left(\frac{-7 \sqrt{2}+4 \sqrt{13}+5 \sqrt{10}}{14}ight)\left(\frac{14 \sqrt{2}-4 \sqrt{13}+5 \sqrt{10}}{14}ight)\left(\frac{14 \sqrt{2}+4 \sqrt{13}-5 \sqrt{10}}{14}ight)}$

Multiplying and solving

$A=\frac{16}{7}$ sq unit

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

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