Question

Find a equation of a line that is perpendicular to y=-4x+3 and passes through the point (4,-1).

Answer 1

Given:

The given equation is:

$y=-4x+3$

A line is perpendicular to the given line and passes through the point (4,-1).

To find:

The equation of required line.

Solution:

The slope intercept form of a line is:

$y=mx+b$

Where, m is slope and b is y-intercept.

We have,

$y=-4x+3$

Here, the slope of the line is -4 and the y-intercept is 3.

Let the slope of required line be m.

We know that the product of slopes of two perpendicular lines is -1. So,

$m\times (-4)=-1$

$m=\dfrac{-1}{-4}$

$m=\dfrac{1}{4}$

The slope of required line is $m=\dfrac{1}{4}$ and it passes through the point (4,-1). So, the equation of the line is:

$y-(-1)=\dfrac{1}{4}(x-4)$

$y+1=\dfrac{1}{4}(x)-\dfrac{4}{4}$

$y=\dfrac{1}{4}x-1-1$

$y=\dfrac{1}{4}x-2$

Therefore, the equation of the required line is $y=\dfrac{1}{4}x-2$.