the line using the point-slope formula y– y 1 =
m(x–x 1 ), where (x 1 ,y 1 ) is the point on the line.
Applying this yields the equation y– 7 = 3(x– 1),
which simplifies to y– 7 = 3x– 3, or equivalently
y= 3x+ 4.
- b.Transforming the equation 3x+ y= 5 into
slope-intercept form simply requires that we
solve for yto obtain the equation y= –3x+ 5.
- b.First, the slope of the line containing the
points (2,3) and (–2,5) is m= = –^24 =
- ^12 . Next, use one of the points, such as (2,3),
and the slope m= –^12 to write the equation of
the line using the point-slope formula y– y 1 =
m(x–x 1 ), where, (x 1 – y 1 ) is the point on the
line. This yields the equation y– 3 = –^12 (x– 2),
which simplifies to y– 3 = –^12 x+ 1, or equiva-
lently y= –^12 x+ 4.
- c.We transform the equationy = – 125 x– ^35
into standard form Ax + By= C, as follows:
y= – 125 x–^35
–15y= 2x+ 9
0 = 2x+ 15y+ 9
2 x+ 15y= –9
- a.We must solve the equation –3y=12x– 3
for y, which can be done by simply dividing
both sides by –3. This yieldsy = –4x+ 1. The
slope of this line is –4.
- d.Solving the equation 6y+ x= 7 for yyields
the equivalent equation y= –^16 x+ ^76 .The
slope of this line is.
- c.We must first determine the equation of the
line. To do so, choose two of the five points
listed and compute the slope. Using (–4,15)
and (–2,11), the slope is m = = –^42 =
–2. Next, use the points, (–4,15), and the slope
m= –2 to write the equation of the line using
the point-slope formula y– y 1 = m(x – x 1 ),
where (x 1 ,y 1 ) is the point on the line. Applying
this yields the equation y– 15 = –2(x– (–4)),
which simplifies to y– 15 = –2x– 8, or equiva-
lently y= –2x+ 7. Now, to determine the miss-
ing value z, we simply substitute x= 2 into this
equation; the resulting value ofyis equal to
the missing value ofz. The substitution yields
y= –2(2) + 7 = 3.
- c.First, the slope of the line containing the
points (0,–1) and (2,3) is m= = ^42 = 2.
Next, use one of the points, such as (2,3), and
the slope m= 2 to write the equation of the line
using the point-slope formulay– y 1 = m(x – x 1 ),
where (x 1 ,y 1 ) is the point on the line. This
yields the equation y– 3 = 2(x– 2), which
simplifies to y– 3 = 2x– 4, or equivalently
y= 2x– 1.
- a.Consider the line whose equation is x= 2. All
points on this line are of the form (2,y), where y
can be any real number. However, in order for
this line to have a y-intercept, at least one of the
points on it must have an x-coordinate of 0,
which is not the case. A vertical line need not
have a y-intercept.
- a.First, we must determine the equation of
the line. The slope of the line is given by m
== ^69 = ^23 . Since the y-intercept of the
line is given to be (0, –6), we conclude that
the equation of the line isy = ^23 x– 6. Now,
observe that substituting the point (–6,–10)
into the equation yields the true statement
–10 = ^23 (–6) – 6. Therefore, the point (–6,–10)
lies on this line.
- d.The slope of a line containing the points
(–3,–1), (0,y), and (3,–9) can be computed
using any two pairs of these points. Specifi-
cally, using (–3,–1) and (3,–9), we see that
0 – (–6)
9 – 0
3 – (–1)
2 – 0
15 – 11
(–4) – (–2)
^1
6
5 – 3
(–2)–2
ANSWERS & EXPLANATIONS–
