Math 135
Linear Equations
Solutions
1. Given is the line with equation y = 3x − 2.
(a) Find five points on the line and arrange them in a table.
Answer 1.
y = 3x − 2
x
y
-2
-8
0
-2
1
1
3
7
10
28
(b) Graph the line.
Answer 2.
y
y = 3x − 2
(1, 1)
x
(0, −2)
(c) Find the x-intercept and the y-intercept.
Answer 3. To find the x-intercept let y = 0 in y = 3x − 2 and solve for x.
Thus, the x-intercept is the point ( 32 , 0).
To find the y-intercept let x = 0 in y = 3x − 2 and solve for y.
Thus, the y-intercept is the point (0, −2).
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Math 135
Linear Equations
Solutions
2. Find the slope-intercept form of the equation of the line through the points (2, 7)
and (5, 2) and graph it.
Answer 4. First calculate the slope. m =
7−2
2−5
= − 53 . So far, we have
5
y =− x+b.
3
To solve for b we substitute the coordinates of a point on the line, for example (2, 7).
Then at the point (2, 7), we have
5
5
31
7 = − · 2 + b ⇐⇒ b = 7 + 2 · ⇐⇒ b =
,
3
3
3
and so the answer is
5
31
y =− x+
.
3
3
We should check our work by verifying that the other point also lies on the line.
In other words, substituting the point (5, 2) we should obtain an identity. Indeed,
31
25 31
6
5
=− +
= =2.
2=− ·5+
3
3
3
3
3
3. Consider the line passing through the point (2, 3) with slope m = −1.
(a) Write down the point-slope equation of the line.
Answer 5.
y − 3 = −(x − 2)
(b) Write the equation in the slope-intercept form.
Answer 6.
y = −x + 5
(c) Find all intercepts.
Answer 7. The x-intercept is the point (5, 0) and the y-intercept is the point
(0, 5).
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Math 135
Linear Equations
Solutions
4. Consider the line y = 2x + 3.
(a) Find the equation in slope-intercept form of a parallel line through (2, 5).
Answer 8. The given line has slope m = 2, so we are looking for a line of the
form y = 2x + b and containing the point (2, 5). Substituting x = 2, it follows
that b = 1 in order for y = 5. Thus, we obtain
y = 2x + 1 .
(b) Find the equation of a perpendicular line through (2, 7).
Answer 9. The line has slope m = 2, so m⊥ = − m1 = − 12 , and a perpendicular
line will have the form y = − 12 x + b. Substituting the point (2, 7) and solving
for b, we obtain
1
y =− x+8.
2
5. Consider the line L given by 2x + 3y = 6.
(a) Find the slope and intercepts of the line.
Answer 10. In the slope-intercept form, we have y = − 32 x + 2 so the slope is
m = − 23 . The x-intercept is the point (3, 0) and the y-intercept is the point (0, 2).
(b) Find a point on the line and a point not on the line.
Answer 11. The point (0, 0) does not lie on the line, but (3, 0) does.
(c) Write the equation of the line in point-slope form.
Answer 12. The slope we already know to be m = − 32 and we can choose the
point (3, 0), so
2
y = − (x − 3) .
3
(d) Find the equation of a line perpendicular to L, but passing through the same
x-intercept as the line L.
Answer 13. We have m = − 23 , so m⊥ = 32 . In slope-intercept form, we have
3
y = x+b,
2
and we need to have this line pass through the point (3, 0). Substituting we
find that b = − 29 , and so the answer is
3
9
y = x− .
2
2
6. Solve:
University of Hawai‘i at Mānoa
y
2x − 5y
14
= 2x − 1
= 10
R Spring - 2014
Math 135
Linear Equations
Solutions
Answer 14. Proceed by elimination: rewrite the system of equations and add them.
We have,
−2x + y = −1
2x − 5y = 10 ,
and whence −4y = 9 ⇒ y = − 49 . Then we substitute y = − 94 into the first equation
and solve for x and obtain x = − 85 . To make sure that (− 58 , − 49 ) is the solution we
check that it also solves the second equation,
5
9
5 45
40
2 −
−5 −
=− +
=
= 10 .
8
4
4
4
4
7. Derive the point-slope form of the equation for a line by following these steps.
(a) Let L be the line passing through the fixed point (x1 , y1 ) and an arbitrary point
(x, y).
(b) Find the general formula for the slope of L.
Answer 15. The slope of L is given by m =
obtain the point-slope form.
y−y1
.
x−x1
Multiplying thru by (x − x1 ), we
8. *Write down a system of 3 linear equations that has
(a) Exactly one solution.
Answer 16. All the above problems have exactly one solution. Take, for example, Problem 6 and introduce a third line which passes through the solution
(− 85 , − 94 ). We use the slope-intercept form with an arbitrary slope, say m = 2,
and obtain

y = 2x − 1

y + 49 = 2(x + 85 ) .

2x − 5y = 10
(b) No solution.
Answer 17. The only three lines in the plane that do not intersect are parallel
lines. We can take for example the line 2x − 5y = 10 and pick 3 different yintercepts.

 2x − 5y = 0
2x − 5y = 5

2x − 5y = 10 .
(c) Infinitely many solutions.
Answer 18. Infinitely many solutions occur when the three lines are in fact the
same line. That is, we have three parallel lines with the same y-intercept.

 2x − 5y = 0
4x − 10y = 0
 2
x − 53 y = 0
3
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Linear Equations
Solutions
9. **Find a pair of points that together with the points (−2, 1) and (2, −2) are the vertices of a square.
Answer 19. Case 1: The segment (−2, 1)(2, −2) is a side of a square.
The line through the points (−2, 1) and (2, −2) has the equation
3
y 0 + 2 = − (x0 − 2) ,
4
and the distance between these points is
p
d = (−2 − 1)2 + (2 + 2)2 = 5 .
Each of the two points we are looking for needs to lie on a line parallel to the one
above and also on a line perpendicular to it and passing thru either (−2, 1) or (2, −2).
That is, we need to solve the systems
y 0 + 2 = 43 (x0 − 2)
p
(y 0 + 2)2 + (x0 − 2)2 = 5 ,
and
y0 − 1 =
p
(y 0 − 1)2 + (x0 + 2)2 =
4
(x0
3
+ 2)
5 .
In the first system, we substitute for (y 0 + 2) in the second equation.
p
(y 0 + 2)2 + (x0 − 2)2 = 5
s
2
4 0
(x − 2) + (x0 − 2)2 = 5
3
r
16 0
(x − 2)2 + (x0 − 2)2 = 5
9
r
25 0
(x − 2)2 = 5
9
25 0
(x − 2)2 = 25
9
(x0 − 2)2 = 9
x02 − 4x0 − 5 = 0
(x0 − 5)(x0 + 1) = 0
(1)
(2)
(3)
(4)
(5)
(6)
(7)
(8)
Thus we obtain the solutions (5, 2) and (−1, −6). Following the same procedure
for the second system, we obtain the solutions (−5, −3) and (1, 5). But our pair of
solutions must lie on a line parallel to the one thru (−2, 1) and (2, −2); i.e., a line
with slope m = − 34 . So, the possible solutions are the pairs of points (1, 5), (5, 2),
and (−5, −3), (−1, −6).
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Linear Equations
Solutions
Case 2: The segment (−2, 1)(2, −2) lies on the diagonal of a square.
In this case we will obtain a smaller square of side length √52 . We need to find the
other diagonal; i.e., the line perpendicular to the segment (−2, 1)(2, −2) and passing
through its midpoint ( −2+2
, 1−2
) = (0, − 21 ). This line has the equation
2
2
y+
4
1
= x,
2
3
and since half the diagonal is 52 , the two points we are looking for need to be distance
5
away from the center of the square, the point (0, − 21 ), as well as the endpoints of
2
(−2, 1)(2, −2). We need to solve the system
(
y 0 + 12
q
(y 0 + 12 )2 + (x0 )2
=
=
4 0
x
3
5 2
2
.
Omitting the algebra, we obtain the points ( 32 , 23 ) and (− 32 , − 52 ), and this is the third
possible solution.
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Math 135
Linear Equations
Solutions
10. ***Find all points such that together with the points (−2, 1) and (2, −2) they are the
vertices of a right triangle.
Answer 20. We have two cases to consider. First, suppose that the line segment
(−2, 1)(2, −2) is the leg of a right triangle. Then, the third vertex lies on a line perpendicular to the line thru (−2, 1) and (2, −2), and passing thru either (−2, 1) or
(2, −2). If (x0 , y 0 ) is the third vertex, then either
4
4
y 0 + 2 = (x0 − 2) or y 0 − 1 = (x0 + 2) .
3
3
The second and more interesting case is that the line segment (−2, 1)(2, −2) is the
hypotenuse of a right triangle. From elementary geometry we recall the Theorem
of Thales, which states that the triangle formed by the diameter of a circle and line
segments joining an arbitrary point on the circle with the endpoints of the diameter
is a right triangle. Thus, we need to find the equation of a circle whose diameter is
the line segment (−2, 1)(2, −2). Using the distance formula, we have the diameter
p
d = (−2 − 1)2 + (2 + 2)2 = 5 ,
and the midpoint of our circle,
1
x1 + x2 y1 + y2
,
= 0, −
.
2
2
2
This is the circle of radius 52 , centered at the point (0, − 21 ).
The answer is disappointing, because we do not explicitly give the coordinates of
a point or points. In fact there are infinitely many possibilities, so listing them
amounts to writing a formula which computes them for us. We have a formula
for the coordinates of every point which solves the problem. If (x0 , y 0 ) is the third
vertex of the right triangle with vertices (−2, 1) and (2, −2), then either
4
y 0 + 2 = (x0 − 2) ,
3
or
4
y 0 − 1 = (x0 + 2) ,
3
or
2 2
1
5
0
(x ) + y +
=
.
2
2
0 2
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