Question

In the following figure, the line passes through the origin and through $P=(1,3)$. Find the equation of a line which passes through $P$ and is perpendicular to the first.

• A. $y=3x+4$
• B. $y=\frac{1}{3}x+4$
• C. $y=\frac{1}{3}x+\frac{10}{3}$
• D. $y=-\frac{1}{3}x+4$
• E. $y=-\frac{1}{3}x+\frac{10}{3}$

Answer 1

The correct option is E $y=-\frac{1}{3}x+\frac{10}{3}$

The line shown in the above figure passes through $(0,0)$ and $(1,3)$.

So the slope of the line $m=\frac{y_2-y_1}{x_2-x_1}=\frac{3-0}{1-0}=\frac{3}{1}=3$

Thus the line which passes through $P(1,3)$ and perpendicular to the first have a negative reciprocal slope which is $-\frac{1}{3}$.

So, the equation of the line is

$\Rightarrow y=mx+b$

$\Rightarrow 3=-\frac{1}{3}\left(1\right)+b$

$\Rightarrow 3=-\frac{1}{3}+b$

$\Rightarrow b=3+\frac{1}{3}$

$\Rightarrow b=\frac{10}{3}$

Put the value to find the equation

$\Rightarrow y=-\frac{1}{3}x+\frac{10}{3}$