Question

Find the distance of the point of intersection of the lines $2x+3y=21$ and $3x-4y+11=0$ from the line $8x+6y+5=0$

Answer 1

since the equation of lines which been intersected is given as;

$2x+3y=21⟶\left(1\right)$

and $3x-4y+11=0\Rightarrow 3x-4y=-11⟶\left(2\right)$

$\left(1\right)\times 4+\left(2\right)\times 3$

$8x+12y=84$

$9x-12y=-33$

$\overline{17x+0=51}$

$x=3\Rightarrow 4y=20\Rightarrow y=5$

$\therefore x=3$ and $y=5$

coordinate of point of intersection is $(3,5)$

since the distance of point $(3,5)$ from the line $8x+6y+5=0$ is given as

$d=\frac{|A{x}_1+B{y}_1+C|}{\sqrt{A^2+B^2}}$

$=\frac{|8\cdot 3+6\cdot 5+5|}{\sqrt{8^2+6^2}}$

$=\frac{59}{10}$