In this question, we have o find the equation of line passing through point (1,1) and is perpendicular to the line
[tex]y = 5 x-1[/tex]
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
[tex]m = - \frac{1}{5}[/tex]
Using slope intercept form, which is
[tex]y=mx+b[/tex]
Substituting the values of x and y from the given point, and the value of m, we will get
[tex]1= - \frac{1}{5} +b[/tex]
adding 1/5 to both sides
[tex]1 + \frac{1}{5} = b \\ b = \frac{6}{5}[/tex]
So the required equation of line is
[tex]y = - \frac{1}{5} x + \frac{6}{5}[/tex]
The equation of the perpendicular line will be: [tex]y=-\frac{1}{5}x+\frac{6}{5}[/tex]
Explanation
Given equation of the line : [tex]y= 5x-1[/tex]
After comparing the above equation with slope intercept form [tex]y= mx+b[/tex], we will get: Slope[tex](m)[/tex] = 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: [tex]-\frac{1}{5}[/tex]
Now, the perpendicular line have slope[tex](m)[/tex] as [tex]-\frac{1}{5}[/tex] and it's passing through the point[tex](x_{1}, y_{1})[/tex] = [tex](1,1)[/tex] , so using the point-slope form......
[tex]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}[/tex]
So, the equation of the perpendicular line will be: [tex]y=-\frac{1}{5}x+\frac{6}{5}[/tex]
