Download Finding Equations of Lines For each part below, use y

Finding Equations of Lines For each part below, use y  mx  b to find the equation of the line. a. a line through the point (4, ‐6) with slope  34 Start by substituting m   34 in the slope‐intercept form to yield y   34 x  b Now substitute the point into the line by setting x  4 and y  6 . This leads to 6   34  4   b
6  3  b
3  b
Using this value for b in the slope‐intercept form above give the line y   34 x  3 . b. a line through the points (‐1, 3) and (2, 6). The slope through the points is m
63
3
 1 2   1 3
Substitute the slope into the slope‐intercept form to give y  1x  b Now take one of the points and substitute it into this equation. Using the ordered pair (2, 6), we can solve for b: 6  1 2   b
4b
This gives us the equation y  1x  4 . c. a line passing through (‐4, 5) and perpendicular to the line with equation 3x  2 y  8 . To find the equation of this line, we must first find the equation of the line 3x  2 y  8 . Solve this equation for y: Finding Equations of Lines 3x  2 y  8
2 y  3x  8 y   32 x  4
The slope of this line is  32 . The slope of a perpendicular line is the negative reciprocal of this number or 23 . Put this number into the slope‐intercept form to yield y  23 x  b Finally, find b by substituting the point into the equation, 5
2
3
 4   b
5   83  b
5  83  b
23
3
b
So the equation of the line is y  23 x  233 . d. a line with x‐intercept of ‐5 and a y‐intercept of 4. These intercepts can be written as ordered pairs (‐5, 0) and (0, 4). The slope between these points is m
40
4
 0   5  5
This leads to the line y  54 x  b Since the y‐intercept is 4, we can substitute it into this line for b to give y  54 x  4 .