Problem of the Week
Problem D and Solution
Slippery Slopes
Problem
Line l1 has equation y = mx + k. Line l1 crosses the y-axis at point P and l2 crosses the x-axis
at point Q. P Q is perpendicular to both line l1 and line l2 . Determine the y-intercept of l2 in
terms of m and k.
l1
y
P
l2
Q
x
Solution
Let the y-intercept of l2 be represented by b.
l1 has equation y = mx + k so we know the slope of l1 is m and the y-intercept is k. Therefore
P , the y-intercept of l1 , is the point (0,k).
Since P Q ⊥ l1 and l2 , it follows that l1 k l2 . Also the slope of P Q is the negative reciprocal of
the slope of l1 . Therefore, slope(P Q) = −1
. Since k is the y-intercept of P Q and the slope of
m
−1
x + k.
P Q is m , the equation of the line through P Q is y = −1
m
The x-intercepts of the line through P Q and the line l2 are the same since both lines intersect
at Q on the x-axis. To find this x-intercept set y = 0 in y = −1
x + k. Then 0 = −1
x + k and
m
m
1
x = k. The result x = mk follows. The x-intercept of the line through P Q and the line l2 is
m
mk and the coordinates of Q are (mk, 0).
We can now find y-intercept of l2 since Q(mk, 0) is on l2 and the slope of l2 is m. Substituting
into the slope-intercept form of the line, y = mx + b, we obtain 0 = (m)(mk) + b which
simplifies to b = −m2 k.
Therefore the y-intercept of l2 is −m2 k.
To the student who solved the problem for l1 with equation y = 4x + 3, you should have
obtained the answer −48 for the y-intercept of l2 . A full solution to this problem is provided
on the next page.
Let l1 be the line y = 4x + 3. From the equation of l1 we know that the slope is 4 and the
y-intercept is 3. Therefore P , the y-intercept of l1 , is the point (0,3).
Since P Q ⊥ l1 and P Q ⊥ l2 , it follows that l1 k l2 . Also, the slope of P Q is the negative
reciprocal of the slope of l1 . Therefore, slope(P Q) = − 14 . Since 3 is the y-intercept of P Q and
the slope of P Q is − 41 , the equation of the line through P Q is y = − 14 x + 3.
The x-intercepts of the line through P Q and the line l2 are the same since both lines intersect
at Q on the x-axis. To find this x-intercept, set y = 0 in y = − 41 x + 3. Then 0 = − 41 x + 3 and
1
x = 3. The result, x = 12, follows. The x-intercept of the line through P Q and the line l2 is
4
12 and point Q is (12, 0).
We can now find equation of l2 since Q(12, 0) is on l2 and the slope of l2 is 4. Substituting
x = 12, y = 0 and m = 4 into y = mx + b, we obtain 0 = (4)(12) + b which simplifies to
b = −48. The equation of l2 is y = 4x − 48 and the y-intercept is −48.
This is the same result we obtained from the general solution on the previous page.