Question

Find the equation of a line perpendicular tox, minus, 3, y, equals, minus, 12x−3y=−12that passes through the point left bracket, 4, comma, minus, 1, right bracket(4,−1).

Answer
Multiple Choice Answers
y, equals, one third, x, plus, 4y=
3
1
​
x+4
y, equals, minus, 3, x, plus, 11y=−3x+11
y, equals, 3, x, plus, 11y=3x+11
y, equals, minus, one third, x, plus, 4y=−
3
1
​
x+4

Answer 1

The equation of the line perpendicular to x−3y=−12 and passing through the point (4, -1) is y=−3x+11.

To find the equation of a line perpendicular to x−3y=−12 and passing through the point (4, -1), we need to determine the slope of the original line and then find the negative reciprocal to get the slope of the perpendicular line.

The given equation x−3y=−12 can be rearranged into the slope-intercept form (y=mx+b) by solving for y: y=1/3x+4

Here, the slope of the original line is 1/3. The negative reciprocal of 1/3 is −3, which is the slope of the perpendicular line.

Now, we use the point-slope form of a linear equation (y−y1)=m(x−x1​) with the given point (4, -1):

y−(−1)=−3(x−4)

y=−3x+11

Therefore, the equation of the perpendicular line is y=−3x+11, and it passes through the point (4, -1).