Find the equation of a line that is perpendicular to 3x-2y=6 and passes through (1,2)
To find the equation of a line that is perpendicular to another line, first, we need to determine the slope of the given line.
The equation 3x - 2y = 6 is in standard form (Ax + By = C), where A = 3, B = -2, and C = 6. To find the slope of this line, we can rearrange the equation in slope-intercept form (y = mx + b) where m is the slope.
Rearranging the equation:
3x - 2y = 6
-2y = -3x + 6
y = (3/2)x - 3
The coefficient of x, 3/2, is the slope of the line. Now, we need to find the slope of the line perpendicular to it.
The slope of a line perpendicular to another line is the negative reciprocal of the slope of the given line. So, the slope of the new line will be -2/3.
Now that we have the slope (-2/3) and a point that lies on the line (1, 2), we can use the point-slope form of a line to find the equation of the line.
The point-slope form of a line is y - y1 = m(x - x1), where (x1, y1) is a point on the line and m is the slope.
Using the values we have, we substitute them into the equation:
y - 2 = (-2/3)(x - 1)
Distribute the slope:
y - 2 = (-2/3)x + 2/3
Rearrange the equation to have y on one side:
y = (-2/3)x + 2/3 + 2
y = (-2/3)x + 8/3
Therefore, the equation of the line that is perpendicular to 3x - 2y = 6 and passes through (1, 2) is y = (-2/3)x + 8/3.