Question

"find the equation of the line passing through the point (1,1) and perpendicular to the line y=5x-1 enter the exact answer. Use improper fractions if it is necessary, do not use mixed fractions."

Answer 1

In this question, we have o find the equation of line passing through point (1,1) and is perpendicular to the line

$y = 5 x-1$

And slope of the given line is 5. And slopes of perpendicular lines are opposite and reciprocal of each other.

So slope of perpendicular line is

$m = - \frac{1}{5}$

Using slope intercept form, which is

$y=mx+b$

Substituting the values of x and y from the given point, and the value of m, we will get

$1= - \frac{1}{5} +b$

adding 1/5 to both sides

$1 + \frac{1}{5} = b \\ b = \frac{6}{5}$

So the required equation of line is

$y = - \frac{1}{5} x + \frac{6}{5}$

Answer 2

The equation of the perpendicular line will be: $y=-\frac{1}{5}x+\frac{6}{5}$

Explanation

Given equation of the line : $y= 5x-1$

After comparing the above equation with slope intercept form $y= mx+b$, we will get:  Slope$(m)$ = 5

As the slope of a perpendicular line is negative reciprocal of the slope of given line, so here the slope of the perpendicular line will be: $-\frac{1}{5}$

Now, the perpendicular line have slope$(m)$ as $-\frac{1}{5}$ and it's passing through the point$(x_{1}, y_{1})$ = $(1,1)$ , so using the point-slope form......

$y- y_{1}= m(x-x_{1})\\ \\ y-1= -\frac{1}{5}(x-1)\\ \\ y-1= -\frac{1}{5}x+\frac{1}{5}\\ \\ y= -\frac{1}{5}x+\frac{1}{5}+1\\ \\ y=-\frac{1}{5}x+\frac{6}{5}$

So, the equation of the perpendicular line will be: $y=-\frac{1}{5}x+\frac{6}{5}$