Question

The line which passes through the intersection point of the lines $x-2y+2=0$ and $x+3y=13$ and parallel to the line $3x+4y=0$ formed a triangle with the co ordinate axes then perimeter of the triangle is-

• A. $34$ unit
• B. $14$ unit
• C. $24$ unit
• D. $48$ unit

Answer 1

The correct option is C $24$ unit

Line passes through
$x-2y+2=0x+3y-13=0x-2y=-2⟶\left(i\right)x+3y=13⟶\left(ii\right)(-)(-)(-)-5y=-15$
$y=3[$Putting value of y, We get $x=4$
$3x+4y=0⟶\left(iii\right)4y=-3xy=\frac{-3}{4}x$
For equation (iii)
When $x=0,y=0x=1y=\frac{-3}{4}x=\frac{-4}{3}y=1$
For equation$⟶\left(i\right)$
When $x=0,y=1x=-2y=0$
For equation$⟶\left(ii\right)$
When $x=0y=\frac{13}{3}x=13,y=0$
As the lines are parallel the stope will be the same which is $y=\frac{-4}{3}x{m}_1=m_2=\frac{4}{3}$
$\therefore$ equation of line passing through $(4,3)$
$(y-3)=\frac{4}{3}(x-4)3y-9=4x-164x-3y-16+9=04x-3y=7$
As equation of x, axis is $y=0$
$x=\frac{7}{4}y=0$
Hence the line touch the x-axis at $(\frac{7}{4},0).$
Applying distance formula an three sides.
$AB=\sqrt{{\left(\frac{7}{4}\right)}^2+{\left(1\right)}^2}=\frac{\sqrt{50}}{2}BC=\sqrt{{\left(\frac{9}{4}\right)}^2+{\left(3\right)}^2}=\frac{\sqrt{90}}{2}CA=\sqrt{16+4}=\sqrt{20}$
Perimeter $\frac{\sqrt{180}}{2}units.$