The equation in slope-intercept form of the line that crossed the x-axis at 36 and is perpendicular to the line represented by y = -4/9x + 5 is ![y=\frac{9}{4} x+36](https://tex.z-dn.net/?f=y%3D%5Cfrac%7B9%7D%7B4%7D%20x%2B36)
<u>Solution:</u>
Given, line equation is ![y = -\frac{4}{9}x + 5](https://tex.z-dn.net/?f=y%20%3D%20-%5Cfrac%7B4%7D%7B9%7Dx%20%2B%205)
We have to find the line equation which is perpendicular to the given line in slope intercept form.
Now, let us find the slope of the given line.
Given line is in slope intercept form,
y = mx + c where m is slope and c is intercept.
So, by comparison, slope of given line is ![\frac{-4}{9}](https://tex.z-dn.net/?f=%5Cfrac%7B-4%7D%7B9%7D)
Now, we know that <em>product of slopes of perpendicular lines = -1
</em>
![\text { Then, slope of perpendicular line } \times \frac{-4}{9}=-1](https://tex.z-dn.net/?f=%5Ctext%20%7B%20Then%2C%20slope%20of%20perpendicular%20line%20%7D%20%5Ctimes%20%5Cfrac%7B-4%7D%7B9%7D%3D-1)
![\text { slope of perpendicular line }=\frac{9}{-4} \times -1=\frac{9}{4}](https://tex.z-dn.net/?f=%5Ctext%20%7B%20slope%20of%20perpendicular%20line%20%7D%3D%5Cfrac%7B9%7D%7B-4%7D%20%5Ctimes%20-1%3D%5Cfrac%7B9%7D%7B4%7D)
And we are also given that, perpendicular line passes x aixs at 36. Which means that intercept of x – axis is 36.
![\text { Now slope intercept form } \rightarrow y=\frac{9}{4} x+36](https://tex.z-dn.net/?f=%5Ctext%20%7B%20Now%20slope%20intercept%20form%20%7D%20%5Crightarrow%20y%3D%5Cfrac%7B9%7D%7B4%7D%20x%2B36)
Thus the equation in slope intercept form is found