Question

Perpendicular distance between 3x-4y+8=0 (1, -1)

Answer 1

Answer:

Step-by-step explanation:

Given line is  $\rm{3x-4y+8=0}$  and given point is  $\rm{(1,-1)}$  ,

We know,

$\rm{The\,\,perpendicular\,\,distance\,\,from\,\,(x_{1},y_{1})\,\,to\,\,the\,\,line\,\,ax+by+c=0\,\,is,}\\\\\rm{d=\dfrac{\left|ax_{1}+by_{1}+c\right|}{\sqrt{{a}^{2}+{b}^{2}}}}$

So, the required distance

$\rm{d=\dfrac{\left|3(1)-4(-1)+8\right|}{\sqrt{{(3)}^{2}+{(-4)}^{2}}}}$

$\rm{\implies\,d=\dfrac{\left|3+4+8\right|}{\sqrt{9+16}}}$

$\rm{\implies\,d=\dfrac{15}{\sqrt{25}}}$

$\rm{\implies\,d=\dfrac{15}{5}}$

$\rm{\implies\,d=3}$