# Solutions for Introduction to Polynomial Calculus

The point-slope form of the equation of a line says that
the rise over the run between

an arbitrary point on a line (x, y) and a particular point (x_{0}, y_{0}) on that
line is constant,

m, called the slope of the line. This describes a relationship of direct
proportionality or

linearity between the rise and the run. The rise is the change in y, y−y_{0}, and
the run is the

change in x, x − x_{0}, so Since the ratio is
undefined for the point (x_{0}, y_{0}), it is

common to cross multiply so that this point fits the equation explicitly: y−y_{0}
= m(x−x_{0}).

If you are given two points on a line, they may be used to compute its slope,
and either

may be used in the point-slope form.

So for (1)-(6) I’m giving not only the slope which the
problem asks for but also the

point-slope equation of the line.

(1) and the equation is y − 1 = 1(x − 0) or y − 2 = 1(x − 1).

(2) and the equation is y − 3 = 2(x − 2) or y − 7 = 2(x − 4).

(3) and the equation is y − 1 = 1/2 (x − 1) or y − 2 = 1/2 (x − 3).

(4) and the equation is y − 4 = −1(x − 1) or y − 2 = −1(x − 3).

(5) and the equation is y−3 = −2/5 (x−(−2)) or y−1 = −2/5 (x−3).

(6) and the equation is y − 0 = 1(x − (−2)) or y − 2 = 1(x − 0).

(7) y − 0 = 2(x − 0)

(8) y − 2 = 5(x − 1)

(9) y − (−1) = −3(x − 2)

(10) y − 1 = 1/2 (x − 1)

(11) y − 5 = −2/3 (x − 0)

(12) y − 0 = 7(x − (−2))

I intentionally prefer the (x−(−a)) form to (x+a) because
it displays the important

information more clearly. I do not require or encourage oversimplification of
answers!

Conversion to slope-intercept form is not required or encouraged either as long
as you

know how to do it. Usually points other than x = 0 are more important and it is
better to

refer equations to the point of interest. The slope-intercept form is nice when
you wish to

extend to polynomials in standard form: but
even polynomials have

useful forms adapted to another point: or
even useful

‘multiple center’ forms:

(13) y = 3x + 1

(14)

(15) Put the equation in slope-intercept form by adding 2y
to both sides, subtracting

4 from both sides, and dividing by 2: y = 3x − 2, so the slope is 3 and the
y−intercept is

−2.

(16) Put the equation in slope-intercept form by
subtracting 2x from both sides, and

dividing by 5: , so the slope is −2/5 and the
y−intercept is 3/5 .

(17) Parallel lines have the same slope, so y − 1 = 3(x − 1)

(18) The equation of any non-vertical line containing the
point (2,−1) is y − (−1) =

m(x − 2). Parallel lines have the same slope, so
So the equation is

y − (−1) = 2(x − 2).

(19) The slope of any line perpendicular to a line with
slope m ≠ 0 is , the

‘negative reciprocal’ rule. So

(20) To find the midpoint of two points and the bisector
of the segment joining them,

compute the simple average their horizontal and vertical coordinates
respectively:

and so the line goes through the point (1,
2). The slope of the segment is

so the slope of any line perpendicular to it is −1/2 and the equation of the
line with this

slope through that point is

(21) The slope of any line perpendicular to a vertical
line x = c is m = 0. So y−1 = 0

or y = 1 whose graph is horizontal.

(22) The equation of any line perpendicular to a
horizontal line y = c is of the form

x = c and its slope is undefined. So x = 2.

(23) The line 2y −x = 4 has slope 1/2 so the equation of a
line through the point (1, 1)

which is perpendicular to this line is y − 1 = −2(x − 1). The intersection of
these lines

may be found by solving the latter for y = −2x + 3 and substituting into the
equation of

the first line: 2(−2x + 3) − x = 4 so x = 2/5 and y = 11/5 . By Pythagoras, this
is the closest

point on the line 2y − x = 4 to the point (1, 1) because the distance to any
other point is

the hypotenuse of a right triangle with one side being the segment between these
points.

This distance is

(24) The line y = 2x− 3 has slope 2 so the equation of a
line through the point (0, 1)

which is perpendicular to this line is The
intersection of these lines

may be found by substituting this into the equation of the first line:
so

x = 8/5 and y = 1/5 . The distance from (0, 1) to this point, hence to the line,
is This distance

is

(25) The point (0, 0) is on the line y = 2x. Both lines
have slope 2 so the equation

of a line through the point (0, 0) which is perpendicular to the line y = 2x + 3
line is

The intersection of those lines may be found
by substituting one

into other: The distance from (0, 0) to this
point,

which is the shortest distance between point on one line and any point on the
other, is