Question

Find the distance of the point (1,1) from the line $12\left( x+6 \right)=5\left( y-2 \right)$ .

Answer 1

Hint : In order to solve this problem we first need to convert the equation in to $Ax+By+C=0$ .
Then the formula for perpendicular distance (d) of a line $Ax+By+C=0$ from a point $O\left( {{x}_{1}},{{y}_{1}} \right)$ is given by, $d=\dfrac{\left| A{{x}_{1}}+B{{y}_{1}}+C \right|}{\sqrt{{{A}^{2}}+{{B}^{2}}}}$ . the distance is irrespective of the sign of the final answer.

Complete step-by-step answer :
We have given the equation of the line and we need to find the distance from that line to another point.
Let the point be (1,1) be $O\left( {{x}_{1}},{{y}_{1}} \right)$ .
The line of the equation given is $12\left( x+6 \right)=5\left( y-2 \right)$ .
We need to solve this equation and write it in the form of $Ax+By+C=0$ .
Solving the equation, we get,
$\begin{aligned} & 12\left( x+6 \right)=5\left( y-2 \right) \\\ & 12x+\left( 12\times 6 \right)-5y+\left( 5\times 2 \right)=0 \\\ & 12x+72-5y+10=0 \\\ & 12x-5y+82=0..............................(i) \\\ \end{aligned}$
Comparing the equation with $Ax+By+C=0$ , we get,
$A=12,B=-5,C=82$ .
The formula for perpendicular distance (d) of a line $Ax+By+C=0$ from a point $O\left( {{x}_{1}},{{y}_{1}} \right)$ is given by,
$d=\dfrac{\left| A{{x}_{1}}+B{{y}_{1}}+C \right|}{\sqrt{{{A}^{2}}+{{B}^{2}}}}..........(ii)$
Therefore, the distance of the point (1,1) is as follows,
$d=\dfrac{\left| \left( 12\times 1 \right)+\left( -5\times 1 \right)+82 \right|}{\sqrt{{{\left( 12 \right)}^{2}}+{{\left( -5 \right)}^{2}}}}$
Solving this we get,
$\begin{aligned} & d=\dfrac{\left| 12-5+82 \right|}{\sqrt{144+25}} \\\ & =\dfrac{89}{13} \end{aligned}$
Hence, the perpendicular distance (d) from the point (1,1) from the line $12\left( x+6 \right)=5\left( y-2 \right)$ is $\dfrac{89}{13}$ units.

Note : We can see that in the numerator there is a modulus sign in order to always consider the positive sign. We need to find the distance therefore; distance is always positive. Also, we are assuming that the distance we are finding is the shortest distance which is the perpendicular distance from the point to the line.