Question

Find the distance of a point $(3,-5)$ from the line $3x-4y-5=0$.

Answer 1

We know that,

Distance (d) of a point $\left(x_2,x_1\right)$ from a line $Ax+By+C=0$ is

$d=\frac{\left|A{x}_1+B{y}_2+C\right|}{\sqrt{A^2+B^2}}$

Now,

$3x-4y-5=0$

The above equation is of the form

$Ax+By+C=0$

Where $A=3,B=-4,C=-5$

& we have to find the distance of the point $(3,-5)$ from the line

So, $x_1=3,y_1=-5$

Now, we know that

$d=\frac{\left|A{x}_1+B{y}_2+C\right|}{\sqrt{A^2+B^2}}$

$\begin{array}{c}=\frac{\left|3\left(3\right)+\left(-4\right)\left(-5\right)-5\right|}{\sqrt{3^2+{\left(-4\right)}^2}}\\ =\frac{\left|9+20-5\right|}{\sqrt{9+16}}\\ =\frac{\left|24\right|}{\sqrt{25}}\\ =\frac{\left|24\right|}{\sqrt{5\times 5}}\\ =\frac{24}{5}\end{array}$

We required distance $\frac{24}{5}$ units.