What is the equation of the line that is perpendicular to y = 7/9x + 15 and goes through the point (1, -2) ?
A. 9x + 7y = 5
B. 9x + 7y = -5
C. 9x + 7y = 23
D. 9x + 7y = -23
To find the equation of a line that is perpendicular to a given line, you need to determine the slope of the given line first.
The given line is y = (7/9)x + 15. The slope of this line is the coefficient of x, which is 7/9.
The slope of a line perpendicular to another line is the negative reciprocal of the slope of the original line. To find the slope of the line perpendicular to the given line, take the negative reciprocal of 7/9: -(9/7).
Now that you have the slope of the perpendicular line, you can use the point-slope form of a line to write the equation. The point-slope form is y - y1 = m(x - x1), where m is the slope and (x1, y1) is a point on the line.
The point given is (1, -2), so you can substitute these values into the equation. The equation becomes y - (-2) = -(9/7)(x - 1).
Simplifying this equation gives y + 2 = -(9/7)(x - 1).
To convert this equation into the standard form (Ax + By = C), multiply both sides of the equation by 7 to get rid of the fraction: 7y + 14 = -9(x - 1).
Distribute the -9 to get -9x + 9: 7y + 14 = -9x + 9.
Move the variables to one side and constants to the other side: 9x + 7y = 5.
Therefore, the equation of the line that is perpendicular to y = (7/9)x + 15 and goes through the point (1, -2) is 9x + 7y = 5.
So, the correct answer is A. 9x + 7y = 5.