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 (x₀, y₀) 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−y₀, and the run is the
change in x, x − x₀, so Since the ratio is undefined for the point (x₀, y₀), it is
common to cross multiply so that this point fits the equation explicitly: y−y₀ = m(x−x₀).
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