Question

the perpendicular distance of a line 4x+3y+5=0 from the point (-1 2) is​

Answer 1

$\underline{\textbf{Given:}}$

$\textsf{Line is 4x+3y+5=0}$

$\textsf{Point is (-1,2)}$

$\underline{\textbf{To find:}}$

$\textsf{The perpendicular distance of a line 4x+3y+5=0 from (-1,2)}$

$\underline{\textbf{Solution:}}$

$\underline{\textbf{Formula used:}}$

$\mathsf{The\;perpendicular\;distance\;from\;(x_1,y_1)\;to\;the\;line\;ax+by+c=0\;is}$

$\mathsf{\left|\dfrac{a\,x_1+b\,y_1+c}{\sqrt{a^2+b^2}}\right|}$

$\textbf{The perpendicular distance of a line 4x+3y+5=0 from (-1,2)}$

$\mathsf{=\left|\dfrac{a\,x_1+b\,y_1+c}{\sqrt{a^2+b^2}}\right|}$

$\mathsf{=\left|\dfrac{4(-1)+3(2)+5}{\sqrt{4^2+3^2}}\right|}$

$\mathsf{=\left|\dfrac{-4+6+5}{\sqrt{16+9}}\right|}$

$\mathsf{=\left|\dfrac{7}{\sqrt{25}}\right|}$

$\mathsf{=\left|\dfrac{7}{5}\right|}$

$\mathsf{=\dfrac{7}{5}\;units}$