Math 8 · A.CED.A.2 · F.LE.A.2

Point-slope form and the line through two points

You know one point on a line and how steep it is, or only two points on it. How do you write its equation?

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When you reach for it

Slope-intercept form, y = mx + b, writes a line from two facts: its slope m, and the number b where it crosses the y-axis. Very often you are not told b.

Point-slope form is the equation that starts from whichever point you happen to have. Reach for it when:

you know the slope of a line and one point on it, any point;
you know two points on a line, since the slope formula turns them into a slope and you are back in the first case;
you have a table of values that changes at a constant rate;
a situation gives two measurements of one quantity that changes steadily with another.

This lesson runs over two days. Day 1 builds point-slope form and two routes from a point and a slope to y = mx + b. Day 2 starts from two points, a table or a situation, and uses a third point as a test.

The procedure

Day 1: point-slope form

Where the form comes from. Suppose a line has slope m and passes through a known point (x_1, y_1). The small numbers are labels, not exponents: x_1 reads “x one” and just means “the first x”, a fixed number. Now let (x, y) stand for any other point of the line. On a straight line the slope between any two of its points is the same, which is what similar slope triangles show. So the slope from (x_1, y_1) to (x, y) is m:

\frac{y - y_1}{x - x_1} = m

Figure 1From the fixed point to any other point of the line, the rise is y - y_1 and the run is x - x_1. Their ratio is always the slope m.

Multiply both sides by x - x_1. This is allowed because x - x_1 is not zero: the point (x, y) is a different point, and the line is not vertical, so its x differs from x_1. The result is

y - y_1 = m(x - x_1)

This is point-slope form. In words: the rise from the fixed point equals the slope times the run from the fixed point. It is the slope formula with one of its two points left general.

Why its solutions are exactly the line. Two things need checking.

First, the fixed point itself satisfies it. Put x = x_1 and y = y_1: the left side is 0 and the right side is m \cdot 0 = 0. The fraction form could not say this, since it would read \displaystyle \frac{0}{0}, which is why the multiplied form is the one used.

Second, nothing off the line satisfies it. Take any solution (x, y). If x = x_1, the right side is 0, so y = y_1 and the solution is the fixed point. If x \ne x_1, divide back by x - x_1: the slope from (x_1, y_1) to (x, y) is m. Only one line passes through (x_1, y_1) with slope m, so (x, y) is on it. Every solution is on the line and every point of the line is a solution: the equation and the line are the same set of points.

Reading the form. Point-slope form subtracts the coordinates of the fixed point, so they appear with their signs flipped. A bracket (x + 3) is (x - (-3)), so the point has x_1 = -3; a left side y + 2 is y - (-2), so y_1 = -2. The slope is the number multiplying the bracket.

Converting to slope-intercept form. Multiply out the bracket by the distributive property, then add y_1 to both sides:

\begin{aligned} y - y_1 &= mx - mx_1 \\ y &= mx + (y_1 - mx_1) \end{aligned}

So the intercept is b = y_1 - m x_1. Multiplying out and adding is quicker than remembering that formula.

Finding b directly. The other route skips point-slope form. Write y = mx + b with the known slope in place of m, put the known point in for x and y, and solve the one-step equation for b. The point is on the line, so its coordinates must satisfy the line’s equation, and that pins down b. Solving y_1 = m x_1 + b gives the same b = y_1 - m x_1, as it must: both routes say “this point is on this line”. Point-slope form keeps the given point visible; substituting is quicker when only y = mx + b is wanted.

Day 2: the line through two points

Two points, slope first. Two different points fix exactly one line, but neither form takes two points directly. So find the slope first with the slope formula,

m = \frac{y_2 - y_1}{x_2 - x_1}

and then use the slope and either point in point-slope form.

Why either point gives the same line. Call the points P and Q. The equation written through P describes the line with slope m through P. That line also passes through Q, because the slope from P to Q is m; that is how m was found. Only one line passes through Q with slope m, so the equation written through Q describes that same line. A line has one point-slope equation for every point on it, and only one slope-intercept equation.

A table. Any two rows of a table are two points. First check that the table really is a line: the slope between neighboring rows must be the same every time. When the steps in x are not equal, compare those quotients, not the changes in y alone. If a row has x = 0, its y is b and can be read off.

A situation. Two measurements are two points (input, output). The slope then has units, output units per input unit, and says how fast the output changes. The intercept b is the output when the input is 0, which may or may not be a moment inside the situation.

A third point as a test. To see whether a point lies on a line, put its x into the equation and compare the result with its y. Any two points lie on some line, so two measurements can never show that a situation is linear. A third one can show that it is not: if it misses the line through the first two, the data do not follow one straight line. The same test is the natural check on any equation you write: the given points must satisfy it.

Worked examples

Example 1

Day 1: a point and a slope

Write the equation of the line with slope 3 through (-5, 4), first in point-slope form and then in slope-intercept form.

Solution

Here m = 3, x_1 = -5 and y_1 = 4. Put them into y - y_1 = m(x - x_1):

y - 4 = 3(x - (-5))

Subtracting -5 is adding 5, so the point-slope form is y - 4 = 3(x + 5).

For slope-intercept form, multiply out the bracket, then add 4 to both sides:

\begin{aligned} y - 4 &= 3x + 15 \\ y &= 3x + 19 \end{aligned}

Check with the given point: at x = -5, 3(-5) + 19 = -15 + 19 = 4, which is its y.

Example 2

Day 1: a fraction slope, by both routes

Write the line with slope \displaystyle -\frac{2}{3} through (9, -5) in slope-intercept form, by converting point-slope form and then by finding b directly.

Solution

Through point-slope form. With x_1 = 9 and y_1 = -5, the left side is y - (-5) = y + 5:

\begin{aligned} y + 5 &= -\frac{2}{3}(x - 9) \\ y + 5 &= -\frac{2}{3}x + 6 \\ y &= -\frac{2}{3}x + 1 \end{aligned}

The bracket multiplied out cleanly because \displaystyle -\frac{2}{3} \cdot (-9) = 6; subtracting 5 from both sides then isolates y.

Finding b directly. Put the slope and the point into y = mx + b:

\begin{aligned} -5 &= -\frac{2}{3}(9) + b \\ -5 &= -6 + b \\ b &= 1 \end{aligned}

Both routes give \displaystyle y = -\frac{2}{3}x + 1, as they must. As a check, take a second point from the answer: x = 3 gives y = -2 + 1 = -1, and the slope from (3, -1) to (9, -5) is \displaystyle \frac{-5 - (-1)}{9 - 3} = \frac{-4}{6} = -\frac{2}{3}.

Example 3

Day 1: reading point-slope form

Which point and which slope does \displaystyle y - 6 = -\frac{1}{2}(x + 8) describe? Write the line in slope-intercept form, find where it crosses the x-axis, and write a second point-slope equation for the same line using that crossing point.

Solution

Rewrite the bracket as a subtraction: x + 8 = x - (-8). So the equation is \displaystyle y - 6 = -\frac{1}{2}(x - (-8)), which names the point (-8, 6) and the slope \displaystyle -\frac{1}{2}.

Multiply out and add 6:

\begin{aligned} y - 6 &= -\frac{1}{2}x - 4 \\ y &= -\frac{1}{2}x + 2 \end{aligned}

The line crosses the x-axis where y = 0: \displaystyle 0 = -\frac{1}{2}x + 2 gives \displaystyle \frac{1}{2}x = 2, so x = 4, the point (4, 0). Point-slope form through it is \displaystyle y - 0 = -\frac{1}{2}(x - 4), that is

y = -\frac{1}{2}(x - 4)

Multiplied out, this is \displaystyle y = -\frac{1}{2}x + 2 again: a second point-slope equation, naming a different point of the same line.

Example 4

Day 2: two points, from either end

Write the equation of the line through (-3, 13) and (5, -11). Do it once from each point.

Solution

The slope comes first, from the slope formula:

m = \frac{-11 - 13}{5 - (-3)} = \frac{-24}{8} = -3

From (-3, 13). The bracket is x - (-3) = x + 3:

\begin{aligned} y - 13 &= -3(x + 3) \\ y &= -3x - 9 + 13 = -3x + 4 \end{aligned}

From (5, -11). The left side is y - (-11) = y + 11:

\begin{aligned} y + 11 &= -3(x - 5) \\ y &= -3x + 15 - 11 = -3x + 4 \end{aligned}

The same equation, y = -3x + 4, as the procedure promised. Each route also checks the other point for free; in the first, x = 5 gives -15 + 4 = -11.

Example 5

Day 2: a line from a table

Show that this table comes from a line, and write its equation.

x	-4	-2	2	6
y	-7	-4	2	8

Solution

The steps in x are 2, 4 and 4, not all equal, so the changes in y (3, 6 and 6) cannot be compared on their own. Compare the slopes between neighboring rows instead:

\frac{3}{2}, \qquad \frac{6}{4} = \frac{3}{2}, \qquad \frac{6}{4} = \frac{3}{2}

They are equal, so every row lies on one line with slope \displaystyle \frac{3}{2}. No row has x = 0, so b cannot be read off. Use the row (2, 2) in point-slope form:

\begin{aligned} y - 2 &= \frac{3}{2}(x - 2) \\ y &= \frac{3}{2}x - 3 + 2 = \frac{3}{2}x - 1 \end{aligned}

Check the other rows: x = -4 gives -6 - 1 = -7, x = -2 gives -3 - 1 = -4, x = 6 gives 9 - 1 = 8. The intercept also fills the gap the table skipped: the missing row is (0, -1).

Example 6

Day 2: draining a hot tub

A pump drains a hot tub at a steady rate. Three minutes after the pump starts, the tub holds 620 liters of water; eight minutes after, it holds 420 liters. Write the volume V, in liters, in terms of the time t, in minutes since the pump started. Say what the slope and the intercept mean, and find when the tub is empty if the rate holds.

Solution

The two measurements are the points (3, 620) and (8, 420), time first because time is the input. The slope is

m = \frac{420 - 620}{8 - 3} = \frac{-200}{5} = -40

in liters per minute, negative because the pump removes 40 liters every minute. Point-slope form through (3, 620), then multiplied out:

\begin{aligned} V - 620 &= -40(t - 3) \\ V &= -40t + 120 + 620 \\ V &= -40t + 740 \end{aligned}

The intercept 740 is the volume at t = 0: the tub held 740 liters when the pump started. Check the other measurement: t = 8 gives -320 + 740 = 420.

Figure 2The hot tub in Example 6. Two measurements fix the line; its ends are the moment the pump starts and the moment the tub is empty.

The tub is empty when V = 0: -40t + 740 = 0 gives 40t = 740, so t = 18.5 minutes. The equation holds only for 0 \le t \le 18.5: before t = 0 the pump was off, and after the tub empties the equation would give negative volumes.

Example 7

Day 2: two quotes and a third

Saul asks a print shop for prices. They quote 57 dollars for 150 flyers and 107 dollars for 400 flyers. Suppose the shop charges a fixed setup fee plus the same price for every flyer. Find the cost C, in dollars, of n flyers and explain both numbers in it. A third quote then arrives: 215 dollars for 1000 flyers. Does it follow the same rule?

Solution

The quotes are the points (150, 57) and (400, 107). The slope is

m = \frac{107 - 57}{400 - 150} = \frac{50}{250} = 0.2

dollars per flyer, so each flyer adds 20 cents. Point-slope form through (150, 57):

\begin{aligned} C - 57 &= 0.2(n - 150) \\ C &= 0.2n - 30 + 57 \\ C &= 0.2n + 27 \end{aligned}

The intercept 27 is the cost of 0 flyers, the setup fee in dollars; the slope 0.2 is the price of each flyer.

Now test the third quote. The rule gives C = 0.2(1000) + 27 = 227 dollars for 1000 flyers, but the shop asked 215. The point (1000, 215) is not on the line; it sits 12 dollars below it. So the three quotes are not linear: the shop prices large orders by a different rule, which the first two quotes alone could never have shown.

Boundaries

Vertical lines have no point-slope form. A vertical line has no slope, so there is no m to write. The vertical line through (x_1, y_1) is x = x_1. The derivation broke at exactly this point: it divided by x - x_1, which is zero for every point of a vertical line.
Horizontal lines are fine. With m = 0, point-slope form reads y - y_1 = 0, that is y = y_1.
Two points must be different. One point repeated does not fix a line: infinitely many lines pass through a single point. Two different points with the same x fix a vertical line, which the slope formula cannot handle, since it would divide by zero.
A table or a situation must change at a constant rate. The method always produces a line through two chosen rows. If the slopes between rows are not all equal, that line misses the other rows, and no line fits the table.
A line from a situation holds only where the situation does. The hot tub’s equation stops at empty; a price rule may change for large orders. The equation does not know its own limits, so they have to be stated.

Where this goes next

Point-slope form is how a line is written whenever the information arrives as “this point, this steepness”. Parallel and perpendicular lines use it straight away: a parallel line through a point keeps the slope and changes the point, and writing such a line is one point-slope equation. Standard form, Ax + By = C, is reached from either form by clearing fractions and gathering x and y on one side. Modeling with linear relationships starts where Day 2 stops: real measurements seldom sit exactly on one line, so the choice becomes which line describes them best and how far its predictions can be trusted.

Further on, the same equation reappears in calculus as the tangent line to a curve, y - f(a) = m(x - a), where the slope m is the steepness of the curve at the point (a, f(a)).

Exercises

Day 1: reading the form

Name the point and the slope, then write the line in slope-intercept form.

y - 9 = 4(x - 2)
\displaystyle y + 3 = -\frac{2}{5}(x + 10)
y = 6(x + 7)

Day 1: writing the line

Write the line in point-slope form, then in slope-intercept form.

Slope 5, through (2, -9)
Slope -4, through (-3, -5)
Slope \displaystyle \frac{3}{4}, through (12, 4)
Slope \displaystyle -\frac{1}{3}, through (-6, 5)
Slope 0, through (8, -11)

Day 1: finding b directly

Find b by putting the point into y = mx + b, then write the line: slope -2, through (7, 3).
Find b the same way for slope \displaystyle \frac{5}{2} through (6, -4), and write the line.
Write y + 1.5 = 0.4(x - 5) in slope-intercept form.

Day 2: the line through two points

Write the line through each pair of points in slope-intercept form.

(-2, -13) and (3, 7)
(-4, -3) and (1, -13)
(-7, -1) and (7, 6)
(3, -6) and (-8, -6)
(-9, 5) and (-1, -1)

Day 2: from a table

A table has x values 1, 3, 5, 7 and matching y values 4, 10, 16, 22. Write the equation of the line it comes from.
Show that the table with x values -6, -2, 4, 10 and matching y values 10, 8, 5, 2 comes from a line, and write its equation.
A table has x values 0, 1, 2, 3 and y values 6, 9, 14, 21. Explain why no line passes through all four rows.

Day 2: from a situation

Under the sea, the total pressure on a diver rises at a steady rate with depth. On Ulysses’s dive it is about 2.5 atmospheres at a depth of 15 meters and about 5.0 atmospheres at 40 meters (an atmosphere is a unit of pressure). Write the pressure P in terms of the depth d, say what the constant term means, and predict the pressure at 65 meters.
Vivienne-Jo’s glider loses height at a steady rate as it comes in to land. Two minutes into the descent it is at 900 meters, and six minutes in it is at 780 meters. Write the height h in terms of the time t in minutes, and find when the glider is at 600 meters.

Day 2: a third point

Write the line through (-5, 10) and (-1, 4). Is (5, -5) on it?
Do (-4, -10), (2, 2) and (6, 11) lie on one line?

Reasoning

For the line through (8, 5) and (11, 11), Tindra writes y - 5 = 2(x - 8) and Wim writes y - 11 = 2(x - 11). Explain, without multiplying out, why both describe the same line. Then confirm it by converting both to slope-intercept form.
Explain why no equation in point-slope form describes the line through (4, -2) and (4, 9), and write an equation that does.

Answers

Point (2, 9), slope 4. Multiply out and add 9: y = 4x - 8 + 9, so y = 4x + 1.
y + 3 = y - (-3) and x + 10 = x - (-10), so the point is (-10, -3) and the slope is \displaystyle -\frac{2}{5}. Multiplying out gives \displaystyle y + 3 = -\frac{2}{5}x - 4, so \displaystyle y = -\frac{2}{5}x - 7.
Read y as y - 0 and x + 7 as x - (-7): the point is (-7, 0), slope 6. Multiplied out, y = 6x + 42.
y + 9 = 5(x - 2). Then y + 9 = 5x - 10, so y = 5x - 19.
y + 5 = -4(x + 3). Then y + 5 = -4x - 12, so y = -4x - 17.
\displaystyle y - 4 = \frac{3}{4}(x - 12). Then \displaystyle y - 4 = \frac{3}{4}x - 9, so \displaystyle y = \frac{3}{4}x - 5.
\displaystyle y - 5 = -\frac{1}{3}(x + 6). Then \displaystyle y - 5 = -\frac{1}{3}x - 2, so \displaystyle y = -\frac{1}{3}x + 3.
y + 11 = 0(x - 8), so y + 11 = 0 and y = -11, a horizontal line.
3 = -2(7) + b = -14 + b, so b = 17 and y = -2x + 17.
\displaystyle -4 = \frac{5}{2}(6) + b = 15 + b, so b = -19 and \displaystyle y = \frac{5}{2}x - 19.
Multiply out: y + 1.5 = 0.4x - 2. Subtract 1.5: y = 0.4x - 3.5.
\displaystyle m = \frac{7 - (-13)}{3 - (-2)} = \frac{20}{5} = 4. Then y - 7 = 4(x - 3) gives y = 4x - 12 + 7, so y = 4x - 5.
\displaystyle m = \frac{-13 - (-3)}{1 - (-4)} = \frac{-10}{5} = -2. Then y + 13 = -2(x - 1) gives y = -2x + 2 - 13, so y = -2x - 11.
\displaystyle m = \frac{6 - (-1)}{7 - (-7)} = \frac{7}{14} = \frac{1}{2}. Then \displaystyle y - 6 = \frac{1}{2}(x - 7) gives \displaystyle y = \frac{1}{2}x - \frac{7}{2} + 6, so \displaystyle y = \frac{1}{2}x + \frac{5}{2}.
\displaystyle m = \frac{-6 - (-6)}{-8 - 3} = \frac{0}{-11} = 0, so the line is horizontal: y = -6.
\displaystyle m = \frac{-1 - 5}{-1 - (-9)} = \frac{-6}{8} = -\frac{3}{4}. Then \displaystyle y + 1 = -\frac{3}{4}(x + 1) gives \displaystyle y = -\frac{3}{4}x - \frac{3}{4} - 1, so \displaystyle y = -\frac{3}{4}x - \frac{7}{4}.
Each step of 2 in x raises y by 6, so the slope is \displaystyle \frac{6}{2} = 3 every time. From (1, 4): y - 4 = 3(x - 1), so y = 3x + 1. Check: x = 7 gives 22.
The slopes between neighboring rows are \displaystyle \frac{8 - 10}{-2 - (-6)} = \frac{-2}{4}, \displaystyle \frac{5 - 8}{4 - (-2)} = \frac{-3}{6} and \displaystyle \frac{2 - 5}{10 - 4} = \frac{-3}{6}, all equal to \displaystyle -\frac{1}{2}, so it is a line. From (-6, 10): \displaystyle y - 10 = -\frac{1}{2}(x + 6), so \displaystyle y = -\frac{1}{2}x - 3 + 10, that is \displaystyle y = -\frac{1}{2}x + 7.
The x steps are all 1, and the slopes between neighboring rows are 3, 5 and 7. A line has the same slope between any two of its points, so these rows are not on one line. The line through the first two rows, y = 3x + 6, gives 12 at x = 2, not 14.
\displaystyle m = \frac{5.0 - 2.5}{40 - 15} = \frac{2.5}{25} = 0.1 atmospheres per meter. Then P - 2.5 = 0.1(d - 15) gives P = 0.1d - 1.5 + 2.5, so P = 0.1d + 1. The constant 1 is the pressure at depth 0, the surface, where only the air presses on the water. At 65 meters, P = 6.5 + 1 = 7.5: about 7.5 atmospheres.
\displaystyle m = \frac{780 - 900}{6 - 2} = \frac{-120}{4} = -30 meters per minute. Then h - 900 = -30(t - 2) gives h = -30t + 60 + 900, so h = -30t + 960. Setting h = 600: 30t = 360, so t = 12 minutes.
\displaystyle m = \frac{4 - 10}{-1 - (-5)} = \frac{-6}{4} = -\frac{3}{2}. Then \displaystyle y - 4 = -\frac{3}{2}(x + 1) gives \displaystyle y = -\frac{3}{2}x - \frac{3}{2} + 4, so \displaystyle y = -\frac{3}{2}x + \frac{5}{2}. At x = 5: \displaystyle -\frac{15}{2} + \frac{5}{2} = -5. Yes, (5, -5) is on the line.
Through the first two points, \displaystyle m = \frac{2 - (-10)}{2 - (-4)} = \frac{12}{6} = 2 and y - 2 = 2(x - 2), so y = 2x - 2. At x = 6 this gives 10, not 11. No, the three points do not lie on one line.
The slope from (8, 5) to (11, 11) is \displaystyle \frac{6}{3} = 2. Tindra’s line has slope 2 through (8, 5), so it also passes through (11, 11). Only one line has slope 2 through (11, 11), and Wim’s is that line, so they are the same. Multiplied out: y = 2x - 16 + 5 and y = 2x - 22 + 11, both y = 2x - 11.
Both points have x = 4, so the line is vertical and the slope formula would divide by 4 - 4 = 0: there is no slope to put in place of m. Every point of the line has x = 4, so the equation is x = 4.