Question

Find the perpendicular distance from (3,8) to the line x+3y-7=0

Answer 1

Answer:

this will be answer of question

Answer 2

The perpendicular  from (3,8) to the line x+3y-7=0 is $2\sqrt 10$ .

Given :

A point ( 3 , 8 ) .

Equation of line x + 3y -7 = 0 .

We know , perpendicular distance from point $(x_1,y_1)$ from line $ax+by+c=0$ is given as :

$D=\dfrac{|ax_1+by_1+c|}{\sqrt{a^2+b^2}}$   ..... ( 1 )

Now , putting all these value in equation 1 .

We get ,

$D=\dfrac{|3+3\times 8-7|}{\sqrt{1^2+3^2}}\\\\D=\dfrac{20}{\sqrt{10}}\\\\D=2\sqrt10$

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Cartesian Coordinate

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