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{align}
  & 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{align} $
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{align}
  & d=\dfrac{\left| 12-5+82 \right|}{\sqrt{144+25}} \\
 & =\dfrac{89}{13}
\end{align} $
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.