SAT Linear Functions

To solve problems about lines in the $xy$-plane on the SAT, you need to know:

• the definition of the slope of a line
• how to find points on a line
• how to find the $x-$ and $y-$intercepts of a line
• the point-slope form of the equation of a line
• the slope-intercept form of the equation of a line
• how to work with intersections of lines
• how to work with parallel and perpendicular lines

Line $m$ is defined by the equation $y_{1}=3x_{1}-6,$ and line $n$ is defined by the equation $y_{2}=-5x_{2}+2.$ If line $m$ intersect line $n$ at point $Q,$ which of the following are point $Q'$s coordinates?

(A) $\ \ (-1, -3)$
(B) $\ \ (1, -3)$
(C) $\ \ (1, 1)$
(D) $\ \ (2, -\frac{2}{5})$
(E) $\ \ (2, -6)$

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Correct Answer: B

Solution 1:

Tip: Draw a picture.
We could find point $Q$ using a graphing calculator, or we could graph the two lines neatly on paper. Point $Q$ is located at $(1,-3),$ as shown below.

Solution 2:

Tip: If two functions intersect at point $(x,y),$ then $f(x) = y = g(x)$.
Let point $Q$ have the coordinates $(x,y).$ Since both lines pass through $(x, y),$ at point $Q:$

line $m$ has the equation $y=3x-6,\qquad (1)$
and line $n$ has the equation $y=-5x+2.\qquad (2)$

Setting these equal to each other, we solve for $x.$

$$\begin{array}{l c l l} 3x-6&=&-5x+2 & \quad \text{set}\ (1)\ \text{equal to}\ (2)\\ 3x&=&-5x+8 &\quad \text{add}\ 6\ \text{to both sides}\\ 8x&=&8 &\quad \text{add}\ 5x\ \text{to both sides}\\ x&=& 1 &\quad \text{divide both sides by}\ 8\\ \end{array}$$

We find $y$ by plugging $x=1$ into $(1)$ or $(2).$ Here we use $(1).$

$$y=3x-6=3\cdot 1-6=3-6=-3$$

So, the point of intersection is $(1, -3).$

Solution 3:

Tip: Plug and check.
Tip: If two functions intersect at point $(x,y),$ then $f(x) = y = g(x)$.
Both lines pass through the same point $Q.$ This means that when $Q$'s coordinates are plugged into the equation for line $m$ we should get a true statement, and when they are plugged into the equation for line $n$ we should also get a true statement.

(A) If $Q=(-1, -3),$ then

$$\begin{array}{ l l c l l l c l} \text{for line}\ m: &y_{1}&=&3x_{1}-6 &\quad \quad \text{for line}\ n: &y_{2}&=&-5x_{2}+2\\ &-3&=&3\cdot(-1)-6 &\quad \quad &-3&=&-5\cdot(-1)+2\\ &-3&\neq&-9 &\quad\quad &-3&\neq&7\\ \end{array}$$

Wrong choice.

(B) If $Q=(1, -3),$ then

$$\begin{array}{ l l c l l l c l} \text{for line}\ m: &y_{1}&=&3x_{1}-6 &\quad \quad \text{for line}\ n: &y_{2}&=&-5x_{2}+2\\ &-3&=&3\cdot(1)-6 &\quad \quad &-3&=&-5\cdot(1)+2\\ &-3&=&-3 &\quad\quad &-3&=&-3\\ \end{array}$$

Both statements are true. Here, we can stop plugging and checking because we found the right answer. For the sake of completion, we show that the remaining choices are wrong.

(C) If $Q=(1, 1),$ then

$$\begin{array}{ l l c l l l c l} \text{for line}\ m: &y_{1}&=&3x_{1}-6 &\quad \quad \text{for line}\ n: &y_{2}&=&-5x_{2}+2\\ &1&=&3\cdot(1)-6 &\quad \quad &1&=&-5\cdot(1)+2\\ &1&\neq&-3 &\quad\quad &1&\neq&-3\\ \end{array}$$

Wrong choice.

(D) If $Q=(2, -\frac{2}{5}),$ then

$$\begin{array}{ l l c l l l c l} \text{for line}\ m: &y_{1}&=&3x_{1}-6 &\quad \quad \text{for line}\ n: &y_{2}&=&-5x_{2}+2\\ &2&=&3\cdot(-\frac{2}{5})-6 &\quad \quad &2&=&-5\cdot(-\frac{2}{5})+2\\ &2&\neq&-7.2 &\quad\quad &2&\neq&4\\ \end{array}$$

Wrong choice.

(E) If $Q=(2, -6),$ then

$$\begin{array}{ l l c l l l c l} \text{for line}\ m: &y_{1}&=&3x_{1}-6 &\quad \quad \text{for line}\ n: &y_{2}&=&-5x_{2}+2\\ &2&=&3\cdot(-6)-6 &\quad \quad &2&=&-5\cdot(-6)+2\\ &2&\neq&-24 &\quad\quad &2&\neq&-28\\ \end{array}$$

Wrong choice.

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Incorrect Choices:

(A)
Tip: Read the entire question carefully.
If you misread the given and solve the problem as if line $m$ is defined by the equation $y_{1}=\boxed{-3x_{1}}-6$ and line $n$ is defined by the equation $y_{2}=\boxed{5x_{2}}+2,$ you will get this wrong answer. Line $m$ is defined by the equation $y_{1}=3x_{1}-6$ and line $n$ is defined by the equation $y_{2}=-5x_{2}+2.$

(C)
Tip: The simplest choice may not be the correct one.
This answer choice is just meant to confuse you. See Solution 3 for how to eliminate it.

(D)
If we set $y_{1}$ equal to $0$ and solve for $x_{1},$ we will get $x_{1}=2.$ Similarly, if we set $y_{2}$ equal to $0$ and solve for $x_{2},$ we will get $x_{2}=-\frac{2}{5}.$ Choice (D) combines these two results into the coordinate point $(2, -\frac{2}{5}),$ and of course it is wrong.

(E)
Tip: Just because a number appears in the question doesn't mean it is the answer.
Solution 3 explains why this answer is wrong.

Line $l$ has a slope of $\frac{2}{3}$ and passes through the point $(-2, -\frac{7}{3})$. Which of the following points is NOT on line $l?$

(A) $\ \ (-3,-3)$
(B) $\ \ (0,-1)$
(C) $\ \ (1,3)$
(D) $\ \ (3,1)$
(E) $\ \ (6,3)$

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Correct Answer: C

Solution 1:

Tip: Draw a picture.
We plot the given points, and expect all of them except for one to lie on a straight line. The exception is the correct answer.

We see from the graph above that choice (C) is the exception.

Solution 2:

Tip: Point-slope form: $y-y_{1}=m(x-x_{1}),$ where $m$ is the line's slope, and $(x_{1}, y_{1})$ is a point on the line.
In this solution, we find the equation of the line using the point-slope form of the equation of a line. In Solution 3, we find it using the slope-intercept form of the equation of a line. In both solutions we will get the same result. Once we find the equation, we test each of the choices, looking for a contradiction.

Let the equation of the line be $y-y_{1}=m(x-x_{1}),$ where (m=-\frac{7}{3}) is its slope, and the point it passes through is $(x_{1}, y_{1})=(-2, -\frac{7}{3}).$

We simplify this:

$$\begin{array}{r c l l l} y-(-\frac{7}{3})&=& \frac{2}{3}(x-(-2))&\quad \text{equation of line}\\ y+\frac{7}{3}&=&\frac{2}{3}x+\frac{4}{3} &\quad \text{simplify}\\ y&=&\frac{2}{3}x-1 &\quad \text{subtract}\ \frac{7}{3}\ \text{from both sides}\\ \end{array}$$

So, the equation of the line is $y=\frac{2}{3}x-1.$

Next, we use the slope-intercept form of the equation of a line. Once we show the result is the same, we test the answer choices for both Solution 2 and Solution 3.

Solution 3:

Tip: Slope-intercept form: $y=mx+b,$ where $m$ is the line's slope, and $b$ its $y$-intercept.
Let the equation for the line be $y=mx+b,$ where $m$ is its slope and $b$ its $y-$intercept.

$$\begin{array}{r c l l l} -\frac{7}{3}&=&\frac{2}{3}\cdot(-2)+b &\quad \text{equation for line}\ l\\ -7 &=& 2\cdot(-2) + 3b &\quad \text{multiply both sides by}\ 3\\ -7 &=& -4 +3b &\quad 2\cdot(-2)=-4\\ -3 &=& 3b &\quad \text{add}\ 4\ \text{to both sides}\\ -1 &=& b &\quad \text{divide both sides by}\ 3\\ \end{array}$$

Again, the equation of the line is $y=\frac{2}{3}x-1.$

For both Solution 2 and Solution 3 we test each answer choice and look for a contradiction.

(A) Testing $(-3,-3)$: $$\begin{array}{r c l} y&=&\frac{2}{3}x-1\\ -3&=&\frac{2}{3}\cdot (-3) -1\\ -3&=&-2-1\\ -3&=&-3\\ \end{array}$$

This is true.

(B) Testing $(0,-1)$: $$\begin{array}{r c l} y&=&\frac{2}{3}x-1\\ -1&=&\frac{2}{3}\cdot (0) -1\\ -1&=&0-1\\ -1&=&-1\\ \end{array}$$

This is true.

(C) Testing $(1,3)$: $$\begin{array}{r c l} y&=&\frac{2}{3}x-1\\ 1&=&\frac{2}{3}\cdot (3) -1\\ -1&=&2-1\\ -3&=&1\\ \end{array}$$

This is false. We found the choice that yields a contradiction. Further testing isn't necessary, but we will verify the two remaining choices.

(D) Testing $(3,1)$:

$$\begin{array}{r c l} y&=&\frac{2}{3}x-1\\ 1&=&\frac{2}{3}\cdot (3)-1\\ 1&=&2-1\\ 1&=&1\\ \end{array}$$

This is true.

(E) Testing $(6,3)$: $$\begin{array}{r c l} y&=&\frac{2}{3}x-1\\ 3&=&\frac{2}{3}\cdot (6) -1\\ 3&=&4-1\\ 3&=&3\\ \end{array}$$

This is true.

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Incorrect Choices:

(A), (B), (D), and (E)
Solution 1 eliminates these choices geometrically. Both solution 2 and Solution 3 eliminate these choices by plugging them into the equation of the line.

In the figure above, line $m$ passes through the points $P(1,0)$ and $Q(0,2).$ What is the value of $x?$

(A) $\ \ -10$
(B) $\ \ -4$
(C) $\ \ 2$
(D) $\ \ 4$
(E) $\ \ 6$

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Solution 1:

Tip: Draw a picture.
Tip: Slope of a line $=\frac{\text{change in}\ y}{\text{change in}\ x}.$

We draw an accurate graph. Starting at point $Q$ we count the change in $y$ and the change in $x$ to get to point $P.$ The change in $y,$ is $2$ tick marks down, and the change in $x,$ is $1$ tick mark to the right.

Starting at point $P(1,0),$ we move $2$ tick marks down, and $1$ to the right, $2$ tick marks down, and $1$ to the right, and again $2$ tick marks down, and $1$ to the right to get to the point with $y$-coordinate equal to $-6$. We check the $x$-coordinate, which if our graph is accurate, should be $4$.

Solution 2:

Tip: The slope of a line is defined as $\frac{\text{change in}\ y}{\text{change in}\ x}.$
Tip: Slope-intercept form: $y=mx+b,$ where $m$ is the line's slope, and $b$ its $y$-intercept.
Let the line be determined by the equation $y=mx+b,$ where $m$ is its slope and $b$ is its $y-$intercept.

Using the two points $P(1,0)$ and $Q(0,2),$ we find the slope of the line.

$\text{slope} = m=\frac{\text{change in}\ y}{\text{change in}\ x} = \frac{2-0}{0-1}=\frac{2}{-1}=-2.$

We already know the $y$-intercept. It is just the point $Q(0,2).$

The equation of the line is $y=-2x+2.$

Finally, we are looking for $x$ when $y=-6.$

$$\begin{array}{r c l l} y&=&-2x+2 &\quad \text{equation of the line}\\ -6&=&-2x+2 &\quad \text{plug in}\ y=-6\\ -8&=&-2x &\quad \text{subtract}\ 2\ \text{from both sides}\\ 4&=&x &\quad \text{divide both sides by}\ -2\\ \end{array}$$

Solution 3:

Tip: If a diagram is drawn to scale, trust it.
The $x$-coordinate should be positive because the point is located in the third quadrant. We can eliminate choices (A) and (B) because they are negative. We know that the distance between the origin and point $Q$ is $2.$ We can divide the plane as shown below, and estimate $x=4.$

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Incorrect Choices:

(A)
Tip: Read the diagram carefully.
If you accidentally plug in $y=6$ instead of $y=-6$ into the equation for the line in Solution 2, you will get this wrong answer.

(B)
Tip: Select the answer with the correct sign!

(C) and (E)
If in Solution 1 you miscount, you will get this wrong answer. If, starting from point $P,$ you trace the slope $2$ times instead of $3,$ you will get choice (C). If you trace the slope $4$ times instead of $3$, you will get choice (E).

Line $m$ has a negative slope and passes through the origin in the $xy$-plane. If line $n$ is parallel to line $m,$ which of the following must be true?

(A) The slopes of lines $m$ and $n$ are negative reciprocals of each other.
(B) Line $n$ has a positive slope.
(C) Line $n$ has a negative slope.
(D) Line $n$ passes through the origin.
(E) Line $n$ has a positive $y$-intercept.

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Correct Answer: C

Solution 1:

Tip: If two lines are parallel, their slopes are equal.
Since line $m$ has a negative slope, line $n$ has the same slope, which is negative. Thus (C) must always be true

Solution 2:

Tip: Draw a picture.
Tip: Look for a counter-example.

We choose a line that has a negative slope and passes through the origin, say $y=-x.$ It is colored in teal in the graph above. We choose another line, line $n$, which is parallel to $y=-x$ and is defined by the equation $y=-x-5.$ Above, it is colored in orange. Line $n$ is a counter-example for each of the wrong choices:

(A) Line $n$ has a slope equal to that of line $m.$ That is, the slope of line $n$ is not the negative reciprocal to that of line $m.$ Eliminate (A). By the way, if the slopes of two lines are negative reciprocals of each other, they are perpendicular. Don't mix up the definitions of parallel and perpendicular lines.
(B) Line $n$ is parallel to line $m,$ yet it has a negative slope. Eliminate (B).
(D) Line $n$ is parallel to line $m,$ but it does not pass through the origin. Eliminate (D).
(E) Line $n$ is parallel to line $m,$ and it has a negative $y$-intercept. Eliminate (E).

The remaining choice is (C), and it is the correct answer.

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Incorrect Choices:

(A), (B), (D), and (E)
Solution 2 gives a counter-example for all of these choices.

If line $p$ is perpendicular to line $n$ in the figure above, what is the slope of line $p?$

(A) $\ \ -5$
(B) $\ \ -\frac{1}{5}$
(C) $\ \ -1$
(D) $\ \ \frac{1}{5}$
(E) $\ \ 5$

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Solution 1:

Tip: If a diagram is drawn to scale, trust it.
Tip: Identify irrelevant information.
Note that line $n$ is irrelevant. We use the given graph to identify the change in $y,$ which is $-5,$ and the change in $x,$ which is $+1.$ This is shown in teal in the graph below.

The slope of line $p$ is $\frac{\text{change in}\ y}{\text{change in}\ x}=\frac{-5}{1}=-5.$

Of course, we could use line $n$ to verify our solution. We've traced the rise and run of line $n$ in orange.

The slope of line $n$ is $\frac{\text{change in}\ y}{\text{change in}\ x}=\frac{1}{5}=\frac{1}{5}.$

The two slopes are negative reciprocals of each other, and therefore the lines are indeed perpendicular.

Solution 2:

Tip: If a diagram is drawn to scale, trust it.
Tip: A line with a negative slope falls from left to right.
First we eliminate choices based on the direction of line $p.$ Then we ignore the sign of line $p$'s slope and compare the changes in $y$ and $x$ for line $p$ to those offered in the choices.

It is evident from the given graph that line $p$ has a negative slope. We can eliminate choices (D) and (E) because they are positive. For the remainder of the analysis, we ignore the negative signs. At any point line $p$ seems to be further away from the $x$-axis than it is from the $y$-axis. This means that the change in $y$ is greater than the change in $x.$

In choice (C), the change in $y$ and the change in $x$ are both equal to $1.$ We can eliminate this answer.

In choice (D), the change in $y$ is $1$ and the change in $x$ is $5.$ So, the change in $y$ is smaller than the change in $x.$ We can eliminate this answer.

In choice (E), the change in $y$ is $5$ and is greater than the change in $x,$ which is $1.$ Therefore, choice (E) is the correct answer. Notice that in this solution, as in Solution 1, the information about line $n$ was irrelevant.

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Incorrect Choices:

(A), (B), (C), and (D)
See Solution 2 for why these choices are wrong.

If you thought these examples difficult and you need to review the material, these links will help:

• Linear Equations
• Slope of a Line
• Finding Points on a Line
• Intercepts of a Line
• Point-Slope Form
• Slope-Intercept Form
• Intersections of Lines
• Parallel and Perpendicular Lines

• The slope of a line is defined as $\frac{\text{change in}\ y}{\text{change in}\ x}.$
• A line with a positive slope rises from left to right.
• A line with a negative slope falls from left to right.
• Slope-intercept form: $y=mx+b,$ where $m$ is the line's slope, and $b$ its $y$-intercept.
• Point-slope form: $y-y_{1}=m(x-x_{1}),$ where $m$ is the line's slope, and $(x_{1}, y_{1})$ is a point on the line.
• The line $x = a$ is a vertical line that crosses the x-axis as $(a,0).$
• The line $y = b$ is a horizontal line that crosses the $y$-axis at $(0,b).$
• If two lines are parallel, their slopes are equal.
• If two lines are perpendicular, their slopes are negative reciprocals of each other.
• If two functions intersect at point $(x,y),$ then $f(x) = y = g(x)$.
• SAT General Tips