Question

the straight line L, has equation y=3x-4 The straight line 1, in perpendicular to L, and passes through the point (9.5) Find an equation of line L

Answer 1

Given:

The equation of given line is

$y=3x-4$

A line is perpendicular to the given line and passes through the point (9,5).

To find:

The equation of the perpendicular line.

Solution:

We have,

$y=3x-4$

On comparing this equation with slope intercept form $y=mx+b$, we get

$m=3$

It means, the slope of the given line is 3.

The product of slopes of two perpendicular lines is -1.

$m\times m_1=-1$

$3\times m_1=-1$

$m_1=-\dfrac{1}{3}$

Point slope form of a line is

$y-y_1=m(x-x_1)$

Where, m is the slope.

The slope of perpendicular line is $-\dfrac{1}{3}$ and it passes through the point (9,5). So the equation of the line is

$y-5=-\dfrac{1}{3}(x-9)$

$y-5=-\dfrac{1}{3}(x)-\dfrac{1}{3}(-9)$

$y-5=-\dfrac{1}{3}x+3$

Adding 5 on both sides, we get

$y-5+5=-\dfrac{1}{3}x+3+5$

$y=-\dfrac{1}{3}x+8$

Therefore, the equation of required line is $y=-\dfrac{1}{3}x+8$.