Question

The distance of the point (1,3) from the line 2x-3y+9=0 measured along a line x-y+1=0 is

• A. $\sqrt{2}$
• B. $\sqrt{5}$
• C. $2\sqrt{2}$
• D. 1

Answer 1

The correct option is C $2\sqrt{2}$

Slope of line $2x-3y+9=0$ is $m_1=\frac{2}{3}$

Slope of line $x-y+1=0$ is $m_2=1$

Now, angle between the two lines is

$tan\theta =\frac{m_2-m_1}{1+m_1{m}_2}=\frac{1-\frac{2}{3}}{1+1\times \frac{2}{3}}=\frac{\frac{1}{3}}{\frac{5}{3}}=\frac{1}{5}$

So, $sin\theta =\frac{1}{\sqrt{26}}$

Now, Perpendicular distance of $(1,3)$ from $2x-3y+9=0$ is

$p=\frac{|A{x}_1+B{y}_1+C|}{\sqrt{A^2+B^2}}$

$=\frac{|2\left(1\right)+(-3)\left(3\right)+9|}{\sqrt{(2{)}^2+(-3{)}^2}}$

$=\frac{|2-9+9|}{\sqrt{4+9}}$

$=\frac{2}{\sqrt{13}}$

Now, distance along the given line $=\frac{p}{sin\theta }=\frac{2}{\sqrt{13}}\times \sqrt{26}=2\sqrt{2}$ units

So, the correct option is $\text{C}$.