Question

Find distance of point $A\left(2,3\right)$ measured parallel to the line $x-y=5$ from the line $2x+y+6=0$.

Answer 1

Step $1$: Find the intersecting point.

Given: Equation of lines,

Equation of line is given as $y=mx+c,m$ is the slope of the line $c$ is the intercept.

$$\begin{gathered} x-y=5\to \left(1\right) \\ 2x+y+6=0\to \left(2\right) \end{gathered}$$

The equation of the line passing through $A\left(2,3\right)$ and parallel to the line $x-y=5$ is,

$$\begin{gathered} \Rightarrow y-3=1\left(x-2\right) \\ \Rightarrow y=x-2+3 \\ \Rightarrow y=x+1\to \left(3\right) \end{gathered}$$

Solving the equation $\left(2\right)$ and $\left(3\right)$, we get the intersection point of lines $\left(2\right)$ and $\left(3\right)$.

$$\begin{gathered} \Rightarrow 2x+y+6=0 \\ \Rightarrow x-y+1=0 \\ \Rightarrow 3x=-7 \\ \Rightarrow x=\frac{-7}{3} \\ \Rightarrow y=\frac{-4}{3} \end{gathered}$$

So, the intersection point is $\left(\frac{-7}{3},\frac{-4}{3}\right)$

Step$2$: Find the distance

Distance formula $=\sqrt{{(x_2-x_1)}^2+{(y_2-y_1)}^2}$

Let $D$ be the required distance,

$$\begin{gathered} \Rightarrow D=\sqrt{{\left(\frac{-7}{3}-2\right)}^2+{\left(\frac{-4}{3}-3\right)}^2} \\ =\sqrt{\frac{{13}^2}{9}+\frac{{13}^2}{9}} \\ =\frac{13}{3}\sqrt{2}units \end{gathered}$$

Hence, the distance is $\frac{13}{3}\sqrt{2}units$ .