Question

What is the equation of the line that is perpendicular to the given line and passes though the point (3,0)? ......What is the answer.....

Answer 1

Answer:  The correct option is (D) $5x-3y=15.$

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

$m=\dfrac{d-b}{c-a}.$

So, the slope of the graphed line is

$m=\dfrac{2-(-1)}{-3-2}\\\\\Rightarrow m=-\dfrac{3}{5}.$

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

$m\times m'=-1\\\\\Rightarrow -\dfrac{3}{5}\times m'=-1\\\\\Rightarrow m'=\dfrac{5}{3}.$

Since the line with slope m' passes through the point (3, 0), so its equation will be

$y-0=m'(x-3)\\\\\\\Rightarrow y=\dfrac{5}{3}(x-3)\\\\\Rightarrow 3y=5x-15\\\\\Rightarrow 5x-3y=15.$

Thus, the required equation of the perpendicular line is $5x-3y=15.$

Option (D) is CORRECT.