Answer: The correct option is (D) ![5x-3y=15.](https://tex.z-dn.net/?f=5x-3y%3D15.)
Step-by-step explanation: We are given to find the equation of a line that is perpendicular to the graphed line and passes through the point (3, 0).
From the graph, we note that
the graphed line passes through the points (2, -1) and (-3, 2).
The SLOPE of a straight line passing through the points (a, b) and (c, d) is given by
![m=\dfrac{d-b}{c-a}.](https://tex.z-dn.net/?f=m%3D%5Cdfrac%7Bd-b%7D%7Bc-a%7D.)
So, the slope of the graphed line is
![m=\dfrac{2-(-1)}{-3-2}\\\\\Rightarrow m=-\dfrac{3}{5}.](https://tex.z-dn.net/?f=m%3D%5Cdfrac%7B2-%28-1%29%7D%7B-3-2%7D%5C%5C%5C%5C%5CRightarrow%20m%3D-%5Cdfrac%7B3%7D%7B5%7D.)
We know that the product of the slopes of two perpendicular lines is -1. So, if m' is the slope of a line perpendicular to the graphed line, then
![m\times m'=-1\\\\\Rightarrow -\dfrac{3}{5}\times m'=-1\\\\\Rightarrow m'=\dfrac{5}{3}.](https://tex.z-dn.net/?f=m%5Ctimes%20m%27%3D-1%5C%5C%5C%5C%5CRightarrow%20-%5Cdfrac%7B3%7D%7B5%7D%5Ctimes%20m%27%3D-1%5C%5C%5C%5C%5CRightarrow%20m%27%3D%5Cdfrac%7B5%7D%7B3%7D.)
Since the line with slope m' passes through the point (3, 0), so its equation will be
![y-0=m'(x-3)\\\\\\\Rightarrow y=\dfrac{5}{3}(x-3)\\\\\Rightarrow 3y=5x-15\\\\\Rightarrow 5x-3y=15.](https://tex.z-dn.net/?f=y-0%3Dm%27%28x-3%29%5C%5C%5C%5C%5C%5C%5CRightarrow%20y%3D%5Cdfrac%7B5%7D%7B3%7D%28x-3%29%5C%5C%5C%5C%5CRightarrow%203y%3D5x-15%5C%5C%5C%5C%5CRightarrow%205x-3y%3D15.)
Thus, the required equation of the perpendicular line is ![5x-3y=15.](https://tex.z-dn.net/?f=5x-3y%3D15.)
Option (D) is CORRECT.