Through (1, -4) and rising 5 units for each unit increase in x.
Its slope must be 5. The equation has the form y = 5x + b. So, —4 = 5(1)+ fe, b = — 9. Thus, the
equation is y = 5* — 9.
Through (5,1) and falling 3 units for each unit increase in x.
Its slope must be -3. So, its equation has the form y = -3x + b. Then, 1 = —3(5) + b, b = 16.
Thus, the equation is y = —3x + 16.
Through the origin and parallel to the line with the equation y = 1.
Since the line y = 1 is horizontal, the desired line must be horizontal. It passes through (0,0), and,
therefore, its equation is y = 0.
Through the origin and perpendicular to the line L with the equation 2x-6y = 5.
The equation of L is y = \x - |. So, the slope of L is 3. Hence, our line has slope —3. Thus, its
equation is y = — 3x.
Through (4,3) and perpendicular to the line with the equation x — l.
The line x = 1 is vertical. So, our line is horizontal. Since it passes through (4,3), its equation is y = 3.
Through the origin and bisecting the angle between the positive *-axis and the positive y-axis.
Its points are equidistant from the positive x- and y-axes. So, (1,1) is on the line, and its slope is 1. Thus,
the equation is y = x.
In Problems 3.61-3.65, find the slopes and y-intercepts of the line given by the indicated equations, and find the
coordinates of a point other than (0, b) on the line.
y = 5x + 4.
From the form of the equation, the slope m = 5 and the y-intercept b = 4. To find another point, set
x = l; then y = 9. So, (1,9) is on the line.
lx - 4y = 8.
y = \x - 2. So, m = J and b = —2. To find another point, set x = 4; then y = 5. Hence, (4,5)
is on the line.
y = 2 - 4x.
m = -4 and b = 2. To find another point, set x = l; then y = -2. So, (1,-2) lies on the line.
y = 2.
m = 0 and b = 2. Another point is (1,2).
y = — f* + 4. So, m = — 5 and b = 4. To find another point on the line, set x = 3; then y = G.
So, (3, 0) is on the line.
In Problems 3.66-3.70, determine whether the given lines are parallel, perpendicular, or neither.
y = 5x - 2 and y = 5x + 3.
Since the lines both have slope 5, they are parallel.
y = x + 3 and y = 2x + 3.
Since the slopes of the lines are 1 and 2, the lines are neither parallel nor perpendicular.
4*-2y = 7 and Wx-5y = l.
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Its slope must be 5. The equation has the form y = 5x + b. So, —4 = 5(1)+ fe, b = — 9. Thus, the
equation is y = 5* — 9.
Through (5,1) and falling 3 units for each unit increase in x.
Its slope must be -3. So, its equation has the form y = -3x + b. Then, 1 = —3(5) + b, b = 16.
Thus, the equation is y = —3x + 16.
Through the origin and parallel to the line with the equation y = 1.
Since the line y = 1 is horizontal, the desired line must be horizontal. It passes through (0,0), and,
therefore, its equation is y = 0.
Through the origin and perpendicular to the line L with the equation 2x-6y = 5.
The equation of L is y = \x - |. So, the slope of L is 3. Hence, our line has slope —3. Thus, its
equation is y = — 3x.
Through (4,3) and perpendicular to the line with the equation x — l.
The line x = 1 is vertical. So, our line is horizontal. Since it passes through (4,3), its equation is y = 3.
Through the origin and bisecting the angle between the positive *-axis and the positive y-axis.
Its points are equidistant from the positive x- and y-axes. So, (1,1) is on the line, and its slope is 1. Thus,
the equation is y = x.
In Problems 3.61-3.65, find the slopes and y-intercepts of the line given by the indicated equations, and find the
coordinates of a point other than (0, b) on the line.
y = 5x + 4.
From the form of the equation, the slope m = 5 and the y-intercept b = 4. To find another point, set
x = l; then y = 9. So, (1,9) is on the line.
lx - 4y = 8.
y = \x - 2. So, m = J and b = —2. To find another point, set x = 4; then y = 5. Hence, (4,5)
is on the line.
y = 2 - 4x.
m = -4 and b = 2. To find another point, set x = l; then y = -2. So, (1,-2) lies on the line.
y = 2.
m = 0 and b = 2. Another point is (1,2).
y = — f* + 4. So, m = — 5 and b = 4. To find another point on the line, set x = 3; then y = G.
So, (3, 0) is on the line.
In Problems 3.66-3.70, determine whether the given lines are parallel, perpendicular, or neither.
y = 5x - 2 and y = 5x + 3.
Since the lines both have slope 5, they are parallel.
y = x + 3 and y = 2x + 3.
Since the slopes of the lines are 1 and 2, the lines are neither parallel nor perpendicular.
4*-2y = 7 and Wx-5y = l.
CHAPTER 3
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3.57
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3.68