Here is the problem. I got the answer but I would like a better understanding.
Find the equation of a line that passes thru(6,-2) and is perpendicular to equation x-3y+16. I got the final answer being -3x+16. Can someone please explain how to get this? I am so confused!
the given line has slope 1/3
so, perpendicular lines will have slope -3
The line you want is thus (using the point-slope form):
y+2 = -3(x-6)
or, if you prefer,
y = -3x + 16
Please Steve, how did you find -3 and 1/3?
To find the equation of a line that is perpendicular to another line, you need to follow these steps:
1. First, rewrite the given equation in the slope-intercept form, which is y = mx + b, where m is the slope and b is the y-intercept.
The given equation is x - 3y + 16 = 0. Rearrange the equation to isolate y:
x - 16 = 3y
y = (1/3)x - (16/3)
From this equation, we can see that the slope of the given line is 1/3.
2. Since the desired line is perpendicular to the given line, the slope of the desired line will be the negative reciprocal of the given line's slope.
The negative reciprocal of 1/3 is -3.
3. Now that we know the slope (-3) and have one point (6, -2) that the line passes through, we can use the point-slope form of a line equation, which is y - y1 = m(x - x1).
Plug in the values we know: (x1, y1) = (6, -2) and m = -3.
y - (-2) = -3(x - 6)
y + 2 = -3x + 18
4. Finally, rearrange this equation to its standard form:
-3x + y = 16
So, the equation of the line that passes through (6, -2) and is perpendicular to the given equation (x - 3y + 16 = 0) is -3x + y = 16.