Answer
(a) $y=-3x+10$
(b) $3x+y-10=0$
(c) Refer to the attached image below for the graph (The broken lines is the graph of $x-3y+16=0$).
Work Step by Step
Write the given equation in slope-intercept form to have:
$x-3y+16=0
\\x+16=3y
\\\dfrac{x+16}{3}=y
\\y=\dfrac{x}{3} + \dfrac{3}{16}$
This equation has a slope of $\dfrac{1}{3}$.
RECALL:
Two lines are perpendicual if their slopes are negative reciprocals of each other (or has a product of -1).
This means that the slope of the line perpendicualr to the line above is $-3$.
Thus, the tentative equation of the perpendicular line is $y=-3x+b$.
To find the value of $b$, substitute the x and y coordinates of $(1, 7)$ into the tentative equation above to have:
$y=-3x+b
\\7 = -3(1) + b
\\7 = -3 + b
\\7+3=b
\\10=b$
Therefore, the equation of the line that passes through (1, 7) and is perpendicular to the line $x-3y+16=0$ is $y=-3x+10$.
(a) $y=-3x=10$
(b) The general equation of a line is $Ax + By + C = 0$, where A, B are not both zero.
Convert the equation of the line to its general form to have:
$y=-3x+10
\\3x+y=10
\\3x+y-10=0$
(c) To graph the line, create a table of values (refer to the attached image below), then plot each point and connect them using a line (refer to the attached image in the answer part).
