What are you supposes to do to write the equation of a line that is perpendicular to the given line and that passes through the given point:
4x-12y=2; (10,-1)
Try the following but check me out.
I would rearrange the equation in the form of y = mx + b.
I believe that's
y = (1/3)x - (1/6).
The line perpendicular to that line has a slope of the negative reciprocal or -3 in this case.
Then use y = mx + b.
y = -3x + b
and substitute the point (10,-1) and solve for b.
I got y=-3x+29. thanks
To write the equation of a line that is perpendicular to the given line and passes through the given point, you can follow these steps:
Step 1: Find the slope of the given line.
- The given line is in the form of Ax + By = C, where A, B, and C are constants.
- Rewrite the equation in slope-intercept form (y = mx + b) by solving for y:
4x - 12y = 2
-12y = -4x + 2
y = (1/3)x - 1/6
- Comparing this equation with y = mx + b, you can see that the slope (m) of the given line is 1/3.
Step 2: Determine the slope of the perpendicular line.
- The slope of a line perpendicular to another line is the negative reciprocal of its slope.
- Take the reciprocal of 1/3, which is 3/1. Then change the sign to negative. Therefore, the slope of the perpendicular line is -3.
Step 3: Use the slope-intercept form and the given point to write the equation.
- You have the slope (-3) and the given point (10, -1).
- Plug the slope and the coordinates of the given point into the point-slope form (y - y1 = m(x - x1)).
y - (-1) = -3(x - 10)
y + 1 = -3x + 30
- Simplify the equation:
y = -3x + 29
Therefore, the equation of the line perpendicular to 4x - 12y = 2 and passing through the point (10, -1) is y = -3x + 29.