For questions 33 and 34, write the equations for the lines parallel and perpendicular to the given line through the given point (one point for each equation).
x - 3y = 9 through (2, -1)
To find the equations for the lines parallel and perpendicular to the given line through the point (2, -1), we need to first determine the slope of the given line.
The given line is in the form: x - 3y = 9
To convert this equation into slope-intercept form (y = mx + b), we need to isolate y:
-3y = -x + 9
Divide both sides by -3:
y = (1/3)x - 3
From this equation, we can see that the slope of the given line is (1/3).
Parallel lines have the same slope, so the equation for the line parallel to the given line would also have a slope of (1/3). We can use the point-slope formula to find the equation:
y - y₁ = m(x - x₁)
Substituting the values of the given point (2, -1) and the slope (1/3), we have:
y - (-1) = (1/3)(x - 2)
y + 1 = (1/3)x - (2/3)
Rearranging this equation, we get the equation for the line parallel to the given line as:
y = (1/3)x - (5/3)
Perpendicular lines have slopes that are negative reciprocals of each other. The negative reciprocal of (1/3) is -3. Hence, the equation for the line perpendicular to the given line would have a slope of -3.
Using the same point-slope formula, we substitute the values of the given point (2, -1) and the slope (-3):
y - (-1) = -3(x - 2)
y + 1 = -3x + 6
Rearranging this equation, we get the equation for the line perpendicular to the given line as:
y = -3x + 5
So, the equations for the lines parallel and perpendicular to the given line through the point (2, -1) are:
Parallel line: y = (1/3)x - (5/3)
Perpendicular line: y = -3x + 5