Question

The distance of the point $\left(1,2\right)$ from the line $x+y+5=0$ measured along the line parallel to $3x-y=7$ is equal to

• A. $4\sqrt{10}$
• B. $40$
• C. $\sqrt{40}$
• D. $10\sqrt{2}$

Answer 1

The correct option is C

$\sqrt{40}$

Explanation for the correct option:

Equation of the line is $3x-y=7$

$y=3x-7$, on comparing with $y=mx+c$ , we know that slope $=3$

The slope of the line parallel to this line will also be $=3$

therefore, Equation of line $AB$:

$\left(y-2\right)=3\left(x-1\right)$

$y=3x-1...\left(1\right)$

We have to find the distance from the line $x+y+5=0...\left(2\right)$

Solving equation $\left(1\right)\mathrm{and}\left(2\right)$:

We get $x=-1,y=-4$

So point $B$ is $\left(-1,-4\right)$

The distance $AB$ is given :$\sqrt{\left(x_2-x_1\right)+\left(y_2-y_1\right)}$

$$\begin{gathered} AB=\sqrt{{\left(-1-1\right)}^2+{\left(-4-2\right)}^2} \\ =\sqrt{40} \end{gathered}$$

Hence, the correct answer is option ( C).