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CHAPTER 3Lines
3.1 Find the slope of the line through the points (—2, 5) and (7,1).
Remember that the slope m of the line through two points (xlt y j and (x2, y2) is given by the equation
Hence, the slope of the given line is
3.2 Find a point-slope equation of the line through the points (1, 3) and (3, 6).
The slope m of the given line is (6 - 3)/(3 - 1) = |. Recall that the point-slope equation of the line through
point (x1, y^) and with slope m is y — yt = tn(x — *,). Hence, one point-slope equation of the given line, using
the point (1, 3), is y — 3 = \(x — 1). Answer
Another point-slope equation, using the point (3,6), is y - 6 = \(x — 3). Answer
3.3 Write a point-slope equation of the line through the points (1,2) and (1,3).
The line through (1,2) and (1,3) is vertical and, therefore, does not have a slope. Thus, there is no
point-slope equation of the line.
3.4 Find a point-slope equation of the line going through the point (1,3) with slope 5.
y -3 = 5(* - 1). Answer
3.5 Find the slope of the line having the equation y - 7 = 2(x - 3) and find a point on the line.
y — 7 = 2(x - 3) is a point-slope equation of the line. Hence, the slope m = 2, and (3, 7) is a point on
the line.
3.6 Find the slope-intercept equation of the line through the points (2,4) and (4,8).
Remember that the slope-intercept equation of a line is y = mx + b, where m is the slope and b is the
y-intercept (that is, the v-coordinate of the point where the line cuts the y-axis). In this case, the slope
m = (8-4)7(4-2) = | = 2 .
Method 1. A point-slope equation of the line is y - 8 = 2(* — 4). This is equivalent to y - 8 = 2* — 8, or,
finally, to y = 2x. Answer
Method 2. The slope-intercept equation has the form y = 2x + b. Since (2,4) lies on the line, we may
substitute 2 for x and 4 for y. So, 4 = 2 - 2 + 6 , and, therefore, b = 0. Hence, the equation is y = 2x.
Answer
3.7 Find the slope-intercept equation of the line through the points (—1,6) and (2,15).
The slope m = (15 -6)/[2- (-1)] = 1 = 3. Hence, the slope-intercept equation looks like y=3x+b.
Since (-1, 6) is on the line, 6 = 3 • (— \) + b, and therefore, b = 9. Hence, the slope-intercept equation is
y = 3x + 9.
3.8 Find the slope-intercept equation of the line through (2, —6) and the origin.
The origin has coordinates (0,0). So, the slope m = (-6 - 0) 1(2 - 0) = -1 = -3. Since the line cuts the
y-axis at (0, 0), the y-intercept b is 0. Hence, the slope-intercept equation is y = -3x.
3.9 Find the slope-intercept equation of the line through (2,5) and (—1, 5).
The line is horizontal. Since it passes through (2,5), an equation for it is y = 5 . But, this is the
slope-intercept equation, since the slope m = 0 and the y-intercept b is 5.
9
, 10 CHAPTER 3
3.10 Find the slope and y-intercept of the line given by the equation 7x + 4y = 8.
If we solve the equation Ix + 4y = 8 for y, we obtain the equation y = — \x + 2, which is the
slope-intercept equation. Hence, the slope m = — I and the y-intercept b = 2.
3.11 Show that every line has an equation of the form Ax + By = C, where A and B are not both 0, and that,
conversely, every such equation is the equation of a line.
If a given line is vertical, it has an equation x = C. In this case, we can let A = 1 and B = 0. If the
given line is not vertical, it has a slope-intercept equation y = mx + b, or, equivalently, — mx + y = b. So,
let A — — m, 5 = 1, and C = b. Conversely, assume that we are given an equation Ax + By = C, with
A and B not both 0. If B = 0, the equation is equivalent to x= CIA, which is the equation of a vertical
line. If B ^ 0, solve the equation for y: This is the slope-intercept equation of the line
with slope and y-intercept
3.12 Find an equation of the line L through (-1,4) and parallel to the line M with the equation 3x + 4y = 2.
Remember that two lines are parallel if and only if their slopes are equal. If we solve 3x + 4y = 2 for y,
namely, y = — f * + i, we obtain the slope-intercept equation for M. Hence, the slope of M is — | and,
therefore, the slope of the parallel line L also is -|. So, L has a slope-intercept equation of the form
y=-\x + b. Since L goes through (-1,4), 4= -\ • (-1) + b, and, therefore, fc=4-i="- Thus, the
equation of L is y = - \x + T •
3.13 Show that the lines parallel to a line Ax + By = C are those lines having equations of the form Ax + By = E
for some E. (Assume that B =£ 0.)
If we solve Ax + By = C for y, we obtain the slope-intercept equation So, the slope is
-A/B. Given a parallel line, it must also have slope —A/B and, therefore, has a slope-intercept equation
which is equivalent to and, thence to Ax + By = bB. Conversely, a line with
equation Ax + By = E must have slope -A/B (obtained by putting the equation in slope-intercept form) and
is, therefore, parallel to the line with equation Ax + By = C.
3.14 Find an equation of the line through (2, 3) and parallel to the line with the equation 4x — 2y = 7.
By Problem 3.13, the required line must have an equation of the form 4x - 2y = E. Since (2, 3) lies on the
line, 4(2) - 2(3) = E. So, £ = 8-6 = 2. Hence, the desired equation is 4x - 2y = 2.
3.15 Find an equation of the line through (2,3) and parallel to the line with the equation y = 5.
Since y = 5 is the equation of a horizontal line, the required parallel line is horizontal. Since it passes
through (2, 3), an equation for it is y = 3.
3.16 Show that any line that is neither vertical nor horizontal and does not pass through the origin has an equation of
the form where b is the y-intercept and a is the ^-intercept (Fig. 3-1).
Fig. 3-1
In Problem 3.11, set CIA = a and CIB = b. Notice that, when y = 0, the equation yields the value
x = a, and, therefore, a is the x-intercept of the line. Similarly for the y-intercept.
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