MATH 147
Lab Key #1
1/26/2016
(1) Find an equation of the line passing through the points (2, 1) and (1, 6) .
Express your final answer in slope-intercept form.
Solution: To write the equation of a line, we need two pieces of information: (1)
a point that the line passes through, and (2) the slope of the line. We have the
first piece of information, but we need to find the second piece of information.
m=
1−6
y2 − y1
=
= −5.
x2 − x1
2−1
Therefore, the slope of the line is -5. Now, we write the equation of the line in
point-slope form:
y − y1 = m ( x − x1 ) .
y − 1 = −5 ( x − 2) .
Finally, we convert to slope-intercept form by isolating y and simplifying:
y − 1 = −5x + 10.
y = −5x + 11.
This is our final answer.
(2) Find an equation of the line passing through the point (1, −6) and parallel
to the line x + 2y = 6. Express your final answer in slope-intercept form.
Solution: Again, we have a point that the line passes through, but we need to
find the slope of our line. To do this, we will first find the slope of x + 2y = 6
by putting it in slope intercept form:
2y = − x + 6.
1
y = − x + 3.
2
The slope of this line is − 12 . Since our line is parallel to this line, our line
has the same slope: m = − 12 . Therefore, plugging into point-slope form [see
problem 1 above]:
1
y − (−6) = − ( x − 1) .
2
1
1
y+6 = − x+ .
2
2
1
1
1
y = − x + − 6.
2
2
Now
1
2
−6 =
1
2
−
12
2
= − 11
2 . Therefore, our final answer is:
1
11
y = − x− .
2
2
(3) Find an equation of the line passing through (−1, −2) and perpendicular
to the line 2x + 5y + 8 = 0. Express your final answer in slope-intercept form.
Solution: Just as in problem 2 above, we will start by finding the slope of the
given line by putting its equation in slope-intercept form:
5y = −2x − 8.
8
2
y = − x− .
5
5
The slope of this line is therefore − 25 . The line in which we are interested is
perpendicular to this line, so taking its negative reciprocal, we see that the
slope of our line is m = 52 . Now using point-slope form, we get the following
equation:
5
y − (−2) = [ x − (−1)] .
2
5
y + 2 = ( x + 1) .
2
5
5
y+2 = x+ .
2
2
5
5
y = x + + 2.
2
2
Since
5
2
+2 =
5
2
+
4
2
= 92 , our final answer is
y=
5
9
x+ .
2
2
2
(4) Find the
exact values of the following expressions:
(a) sin 3π
4 (b) tan
4π
3
Solution:
(a) First, we draw our unit circle. Then we draw where our angle is on the
circle, as in the figure below. From this angle, we draw a right triangle. This is
one of our special right triangles, and so we mark out the sides. Then
1
opp.
3π
= √ .
sin
=
4
hyp.
2
(b) Now we do the same thing with
4π
3 .
Now, we can see that
√ − 23
√
4π
opp.
= 3.
tan
=
=
3
adj.
− 12
3
(5) Find all values of x ∈ [0, 2π ) such that
√
cos x = −
3
.
2
Solution: We are looking for angles with a negative cosine. Since cosine corresponds to the x-coordinate, we are looking
for angles in Quadrants II and III.
√
The reference angle giving a cosine of 23 is π6 . Hence, the values of x that solve
the equation are the angles in Quadrants II or III with reference angle π6 . These
7π
are 5π
6 and 6 .
5π
7π
∴x=
or
.
6
6
4