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

Engineering

Mathematics

Line of Greatest Slope

Question

\sqrt{2}

\sqrt{5}

2 \sqrt{2}

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Slope of line

2 x - 3 y + 9 = 0

m_{1} = \frac{2}{3}

Slope of line

x - y + 1 = 0

m_{2} = 1

Now, angle between the two lines is

t a n θ = \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,

s i n θ = \frac{1}{\sqrt{26}}

Now, Perpendicular distance of

(1, 3)

from

2 x - 3 y + 9 = 0

p = \frac{∣ A x_{1} + B y_{1} + C ∣}{\sqrt{A^{2} + B^{2}}}

= \frac{∣ 2 (1) + (- 3) (3) + 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}{s i n θ} = \frac{2}{\sqrt{13}} \times \sqrt{26} = 2 \sqrt{2}

units

So, the correct option is

C

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