Finding the Equation of a Line: Free Response

Point-slope form is $y - y_1 = m(x - x_1)$, so write the template down first and only then decide what each letter is. Here the point supplies $x_1 = 6$ and $y_1 = -4$, and the slope supplies $m = -\tfrac{1}{2}$. Substituting all three at once:

$y - (-4) = -\tfrac{1}{2}(x - 6).$

The left side subtracts a negative number, which is an addition, so the equation tidies to

$y + 4 = -\tfrac{1}{2}(x - 6).$

The right side keeps its minus sign untouched, because $x_1 = 6$ is already positive. As a check, put the point itself back in: the left side is $-4 + 4 = 0$ and the right side is $-\tfrac{1}{2}(6 - 6) = 0$, so $(6, -4)$ does lie on it.

Look at the two coordinate pairs before reaching for any formula. They differ in their second coordinate but agree in their first, so both points sit in the same vertical column and the line through them runs straight up and down.

The slope formula would ask for the difference of the two first coordinates on the bottom, and that difference is $-5 - (-5) = 0$, so there is no slope to compute and nothing for the $m$ in point-slope form to receive. None of that is needed. Every point of the line shares the same first coordinate, and an equation can say exactly that:

$x = -5.$

Both given points satisfy it, since each has first coordinate $-5$, and no point outside that column does.

Read the template as a list of what it needs: $y - y_1 = m(x - x_1)$ asks for a point, which every line has plenty of, and for a number $m$, which not every line owns.

A line that is not vertical has a slope, so both ingredients exist and the form can be written down. Running that the other way, any equation of the shape $y - y_1 = m(x - x_1)$ describes a line of slope $m$, and a line with a slope is not vertical. So the lines point-slope form can express are exactly the non-vertical ones, with nothing left over on either side.

A vertical line is the single exception. Its slope is undefined, not zero, so there is no number to put in the $m$ position at all, and the form never gets started. Such a line is written directly as $x = c$ instead, as in part B.

A slope of $0$ is a different situation entirely, and the form handles it without complaint. Substituting $m = 0$ makes the whole right side vanish:

$y - y_1 = 0(x - x_1), \qquad y - y_1 = 0, \qquad y = y_1.$

That is the horizontal line through the point, whose second coordinate never changes. So zero slope is a case the form covers, and undefined slope is a case it cannot reach; the two are opposite ends of the situation, not the same one.

Decide first which quantity plays the horizontal role. Time is what is being read off the clock, so let $t$ run horizontally and $V$ vertically. Each reading then becomes an ordered pair: at $3$ minutes, $(3, 26)$; at $8$ minutes, $(8, 46)$.

The rate is steady, so the same amount arrives in every minute, and that is precisely what a constant slope means. Measure it from the two readings:

$m = \frac{46 - 26}{8 - 3} = \frac{20}{5} = 4.$

Its units are liters per minute, since a difference of liters was divided by a difference of minutes. Now feed that slope and one of the two readings into point-slope form. Taking $(3, 26)$:

$V - 26 = 4(t - 3).$

Using the other reading instead would give $V - 46 = 4(t - 8)$, the same line written from a different starting point.

Standard form allows no brackets and no fractions, so distribute first:

$V - 26 = 4t - 12.$

Now collect the two variables on one side and the constants on the other. Subtracting $4t$ from both sides and adding $26$ to both sides gives

$-4t + V = 14.$

The coefficient in front of $t$ is negative, and the convention asks for it to be zero or positive, so multiply the whole equation through by $-1$:

$4t - V = -14.$

The coefficients are integers and share no common factor. Check it against the reading it was built from: at $t = 3$ and $V = 26$, the left side is $12 - 26 = -14$, which matches.

The equation is a rule that every pair of matching readings obeys, so fix the volume at $90$ and solve for the time. Using the standard form:

$4t - 90 = -14, \qquad 4t = 76, \qquad t = 19.$

The units come from what $t$ was defined to count, so the answer is $19$ minutes after filling began.

Check it against the point-slope version, which is an independent route to the same claim:

$V - 26 = 4(19 - 3) = 64, \qquad V = 90.$

A sanity check on the size: $90$ liters is more than the $46$ liters recorded at $8$ minutes, so the time must fall later than $8$ minutes, and $19$ does.

Do the two substitutions first. At $(3, 26)$ and then at $(8, 46)$:

$4(3) - 26 = 12 - 26 = -14, \qquad 4(8) - 46 = 32 - 46 = -14.$

Both match the right-hand side, so both readings lie on the line.

Now the reason the second one is worth doing. Point-slope form is built by forcing the equation through the chosen point: substituting that point makes both sides zero no matter what number was written in the $m$ position. So testing it confirms the substitution was copied correctly and nothing else.

Suppose the rate had been miscomputed as $5$ instead of $4$. Point-slope form with the same first reading would give $V - 26 = 5(t - 3)$, and the first reading still passes, because it is built in. The second reading is what exposes the error:

$V = 26 + 5(8 - 3) = 51 \neq 46.$

So the reading held back from the construction is the only one carrying independent information about the rate. That is why a two-point problem should always be checked against both of its points.

An equation of a line is the rule its points obey: a pair $(x, y)$ satisfies the equation exactly when the point sits on the line. So any point the line is required to pass through must satisfy any correct equation for it, and that gives a test costing one substitution and no rederivation.

The line is required to pass through $(-3, 4)$. Substituting:

$2(-3) - 3(4) = -6 - 12 = -18.$

The right-hand side is $-6$, and $-18 \neq -6$, so the pair $(-3, 4)$ does not satisfy Rowan's equation. That means the point does not lie on the line Rowan's equation describes, and the line described was required to pass through it. One failure is enough, because a correct equation would have to pass every such test, not most of them.

Notice what this test does not tell you: it says the answer is wrong without saying where it went wrong. That is the next part's job.

Substitute into $y - y_1 = m(x - x_1)$ with $x_1 = -3$, $y_1 = 4$ and $m = \tfrac{2}{3}$. The first coordinate is negative, so the bracket subtracts a negative number and becomes an addition:

$y - 4 = \tfrac{2}{3}\bigl(x - (-3)\bigr) = \tfrac{2}{3}(x + 3).$

The second coordinate is positive, so the left side keeps its minus sign. To reach standard form, clear the fraction by multiplying both sides by $3$, then distribute:

$3y - 12 = 2(x + 3) = 2x + 6.$

Collect both variables on the left and the constants on the right, then multiply through by $-1$ so the coefficient of $x$ is positive:

$-2x + 3y = 18, \qquad 2x - 3y = -18.$

Check: $2(-3) - 3(4) = -6 - 12 = -18$, matching the right-hand side, so the given point does lie on this one.

Do not compare the two finished equations and guess. Start from a single wrong number in the template and push it through every step the note describes; if it lands on what was handed in, the diagnosis accounts for the answer exactly, and if it does not, the diagnosis is wrong.

The template needs three numbers. The slope $\tfrac{2}{3}$ appears untouched in the handed-in equation, since clearing a denominator of $3$ is what produces the coefficients $2$ and $3$. The second coordinate $y_1 = 4$ is positive, so it offers no sign to lose. That leaves the first coordinate, and it is the one that is negative.

Suppose Rowan wrote $x - 3$ in the bracket, as though $x_1$ were $3$ rather than $-3$. Following the note's own steps, substitute:

$y - 4 = \tfrac{2}{3}(x - 3).$

Multiply by $3$ and distribute:

$3y - 12 = 2x - 6.$

Collect and fix the sign convention:

$-2x + 3y = 6, \qquad 2x - 3y = -6.$

That is the equation Rowan handed in, character for character, so the single slip accounts for the whole of it. Everything else Rowan did was correct: the slope was placed correctly, the fraction was cleared correctly, and the terms were collected correctly.

One more thing the slip explains. Both the correct answer and Rowan's have the same left-hand side, $2x - 3y$, and differ only in the constant. A sign lost from $x_1$ inside the bracket can only ever move the constant, never the coefficients, which is why the two equations look so nearly alike.

Two points, so measure the direction between them first:

$m = \frac{1 - 5}{6 - (-2)} = \frac{-4}{8} = -\tfrac{1}{2}.$

Feed that slope and either point into point-slope form. Taking $P(-2, 5)$, and remembering that subtracting the negative first coordinate turns into an addition:

$y - 5 = -\tfrac{1}{2}(x + 2).$

For standard form, multiply both sides by $2$ to clear the fraction, distribute, and collect:

$2y - 10 = -(x + 2) = -x - 2, \qquad x + 2y = 8.$

Check both plotted points, since a wrong slope can still satisfy the one you built from. At $P$: $-2 + 10 = 8$. At $Q$: $6 + 2 = 8$. Both match.

The first step of part A was the slope, so try it and watch where it fails:

$\frac{1 - 5}{-2 - (-2)} = \frac{-4}{0}.$

Division by zero produces no number at all, so the slope is undefined. That is a stronger statement than saying the slope is $0$: a slope of $0$ is a perfectly good number belonging to a horizontal line, whereas here there is no number to have. And with no slope, point-slope form has nothing to put in its $m$ position, so the method cannot even be written down, let alone finished.

The line still exists, and it can be described without measuring any direction. The two points agree in their first coordinate, so they sit one above the other and the line through them runs straight up and down. Every point of that line has first coordinate $-2$, and the equation says exactly that:

$x = -2.$

Both given points satisfy it, since each has first coordinate $-2$, and the second coordinate is left completely free, which is right: the line contains a point at every height.

Compare the two templates by what letters they contain.

Point-slope form, $y - y_1 = m(x - x_1)$, contains a slope. So it can be written for a line only if that line has one, which rules out the vertical lines and nothing else. Running the statement the other way, every equation of that shape describes a line with slope $m$, and a line with a slope is not vertical. Both directions hold, so the lines point-slope form expresses are exactly the non-vertical ones.

Standard form, $Ax + By = C$, contains no slope. It has only coefficients, and a coefficient is allowed to be zero as long as $A$ and $B$ are not both zero at once. That extra freedom is what lets it reach the case point-slope form cannot: setting $B = 0$ leaves an equation in $x$ alone. The part B line is written this way,

$1x + 0y = -2,$

which is the equation $x = -2$ with the missing variable shown explicitly. The horizontal case is the mirror image, with $A = 0$.

Every other line, being non-vertical, has a point-slope equation, and clearing its fraction and collecting the terms always produces an $Ax + By = C$. So standard form expresses every line, vertical ones included, which is exactly why the lesson calls it the natural way to write a vertical line.

The price is the one the lesson also names: standard form hides the slope and the point from view, while point-slope form displays both at a glance. Neither form is better; they buy different things, and the vertical line is where their reach comes apart.

Measure the slope once, from the two points:

$m = \frac{4 - 2}{3 - (-1)} = \frac{2}{4} = \tfrac{1}{2}.$

That one number serves both versions, because a line has only one slope. Substituting each point in turn:

$y - 2 = \tfrac{1}{2}(x + 1), \qquad y - 4 = \tfrac{1}{2}(x - 3).$

The two look nothing alike. Convert the first by multiplying through by $2$ and collecting:

$2y - 4 = x + 1, \qquad x - 2y = -5.$

And the second the same way:

$2y - 8 = x - 3, \qquad x - 2y = -5.$

The two starting points led to the same standard form, so the choice of starting point changed only how the equation looked before it was tidied, never which line it described. Check both given points against it: $-1 - 4 = -5$ and $3 - 8 = -5$.

Two tests are available, and it is worth using both, since they fail in different ways.

Candidate (i), $3x - 6y = -15$. Every coefficient is three times the corresponding one in $x - 2y = -5$, so dividing through by $3$ reproduces part A's equation exactly. Substitution agrees: at $(-1, 2)$, $-3 - 12 = -15$; at $(3, 4)$, $9 - 24 = -15$. Both of $L$'s points satisfy it, so it describes $L$.

Candidate (ii), $x - 2y = 5$. One substitution settles it:

$(-1) - 2(2) = -5 \neq 5.$

The point $(-1, 2)$ lies on $L$ but does not satisfy this equation, so this is not $L$. Notice how little the equation's appearance had to change: only the constant differs from part A's, and that was enough.

Candidate (iii), $y + 2 = \tfrac{1}{2}(x - 1)$. It is in point-slope form, built from the point $(1, -2)$, so bring it to the same shape as the others by multiplying through by $2$ and collecting:

$2y + 4 = x - 1, \qquad x - 2y = 5.$

That is candidate (ii) written differently, so it fails for the same reason. Substituting directly confirms it: $(-1) - 2(2) = -5$, and not $5$.

So of the three, only (i) is $L$. Two of them are the same line as each other, and it is not $L$: looking different is no evidence at all, in either direction.

Start from what it means for a pair to satisfy an equation: substituting it makes the two sides equal numbers.

Suppose $(x_0, y_0)$ satisfies $Ax + By = C$, so that $Ax_0 + By_0 = C$ is a true numerical statement. Multiplying both sides of a true statement by the same number keeps it true:

$kAx_0 + kBy_0 = kC.$

So the pair satisfies the scaled equation too. Now run it backwards. If a pair satisfies $kAx + kBy = kC$, divide both sides by $k$, which is allowed precisely because $k$ is not zero, and the original equation comes back. So the pair satisfies that one as well.

Both directions hold, so the two equations are satisfied by exactly the same pairs. The line is nothing more than the set of pairs that satisfy the equation, so the two equations describe the same line. The reversibility is the whole argument, and it is where $k \neq 0$ earns its place: multiplying by $0$ turns every equation into $0 = 0$, which every pair satisfies, and that step cannot be undone.

The consequence is that a single line owns an entire family of standard-form equations, all scalings of one another:

$x - 2y = -5, \qquad 2x - 4y = -10, \qquad -3x + 6y = 15.$

That is inconvenient, because comparing two lines by comparing their coefficients only works if each line has one agreed equation. The conventions supply exactly that. Dividing out any common factor rules out $2x - 4y = -10$, and requiring the coefficient of $x$ to be zero or positive rules out $-3x + 6y = 15$.

Those two rules settle every line whose equation really contains an $x$ term, and one kind of line does not. A horizontal line has $A = 0$, and then $A \ge 0$ holds whatever else is done, so it rules nothing out. The line $y = 1$ can be written $0x + y = 1$ or $0x - y = -1$, and both have whole-number coefficients with no common factor and a coefficient of $x$ that is not negative, yet they are one line written twice. So the convention carries a third clause: when $A = 0$, the coefficient of $y$ must be positive. That keeps $0x + y = 1$ and rules out $0x - y = -1$, and with it exactly one equation of the family survives, for every line.

So the conventions are not decoration: they are what makes two tidied equations comparable at a glance, and they are why part B's first candidate had to be reduced before its coefficients meant anything.