Graphing Lines: Chapter Test

Solution

Correct answer: B

You chose A: Divides the run by the rise, reporting $\tfrac{4}{8}$ in place of $\tfrac{8}{4}$.

You chose C: Stops at the rise $10 - 2$ and never divides by the run.

You chose D: Subtracts the $y$-values in one order and the $x$-values in the other, computing $\tfrac{10 - 2}{-3 - 1}$.

Solution

Correct answer: D

You chose A: Sets $x = 0$ rather than $y = 0$, which locates the crossing of the other axis.

You chose B: Sets $y = 0$ correctly but then divides $40$ by $-4$, the coefficient of the term that was just zeroed out.

You chose C: Solves for $x$ correctly but writes the value in the second slot of the ordered pair.

Solution

Correct answer: A

You chose B: Takes the first number on the right as the slope, though it is the constant term.

You chose C: Takes the coefficient of $x$ but leaves behind the minus sign attached to it.

You chose D: Takes the leading number as the slope and moves the $x$-term's minus sign onto it.

Solution

Correct answer: C

You chose A: Swaps the flat case for the upright one; $y = c$ locks the height, so its rise is $0$ and its slope is $0$.

You chose B: Reads an equation with no constant term as one with no slope, though its slope is the coefficient $6$.

You chose D: Reads an equation carrying both letters as one with no single slope, though solving for $y$ gives the slope $-1$.

Solution

Correct answer: A

You chose B: Substitutes the coordinates in reverse, computing $3(4) - 2(-3)$ for the pair $(-3, 4)$ and accepting the $18$ it produces.

You chose C: Sets $x = 0$ and then solves $-2y = 18$ as though the coefficient were positive.

You chose D: Sets $y = 0$ and divides $18$ by $2$, the coefficient of the term just zeroed out, instead of by $3$.

Solution

Correct answer: C

You chose A: Changes the sign without flipping the fraction, so the product of the slopes is $-\tfrac{9}{25}$ rather than $-1$.

You chose B: Flips the fraction without changing the sign, so the product of the slopes is $+1$ rather than $-1$.

You chose D: Gives the slope of a parallel line, which copies the slope instead of reversing the direction.

Solution

Correct answer: D

You chose A: Drops the minus sign on each coordinate, subtracting $7$ and $2$ instead of $-7$ and $-2$, so both signs come out wrong.

You chose B: Subtracts the negative $y$-coordinate correctly but treats $x_1$ as $+2$, so the parenthesis subtracts where it should add.

You chose C: Subtracts the negative $x$-coordinate correctly but treats $y_1$ as $+7$, so the left side subtracts where it should add.

Solution

Correct answer: B

You chose A: Swaps the two cases, reading a repeated $y$-value as the upright line and a repeated $x$-value as the flat one.

You chose C: Reports $\tfrac{13}{0}$ as $0$, treating a zero denominator the way a zero numerator behaves.

You chose D: Calls any slope built from a repeated coordinate undefined, without checking whether the repetition lands in the rise or in the run.

Solution

Correct answer: C

You chose A: Reads both numbers off the standard-form equation without first solving for $y$.

You chose B: Divides the $x$-term by $4$ but leaves the constant on the right undivided.

You chose D: Divides every term correctly but drops the minus sign that carrying $6x$ across the equals sign produced.

Solution

Correct answer: A

You chose B: Reports the constant term, which is the depth at $t = 0$, as though it answered a question about time.

You chose C: Reports the $6$ that multiplies $t$, the metres of depth lost each minute, instead of solving for $d = 0$.

You chose D: Carries the $42$ across the equals sign without changing its sign, solving $6t = -42$.

Solution

Correct answer: D

You chose A: Matches the two slopes by size alone and overlooks that one of them is negative.

You chose B: Treats a change of sign as the perpendicular condition, though the fraction must turn over as well.

You chose C: Sees $\tfrac{2}{7}$ in both and concludes they are one line, without noticing the difference in sign or comparing the constants $9$ and $2$.

Solution

Correct answer: B

You chose A: Subtracts the $y$-values in one order and the $x$-values in the other, so the slope arrives as $+3$, and then fits the constant to $(-1, 8)$.

You chose C: Finds the slope correctly but takes the first point's $y$-coordinate as the constant, as though $(-1, 8)$ sat on the $y$-axis.

You chose D: Finds the slope correctly but solves $-4 = -9 + b$ in the wrong direction, reporting $b = -9 - (-4)$.

Solution

Correct answer: A

You chose B: Fixes $k = 2$ by matching it to the coefficient of $x$ rather than by using the point that was given.

You chose C: Finds $k$ correctly but then sets $y = 0$, which locates the crossing of the other axis.

You chose D: Reads the constant on the right as the height of the crossing, without dividing by the coefficient of $y$.

Solution

Correct answer: D

You chose A: Loses the minus sign on the slope, stepping $4$ up from $(-2, 7)$ instead of $4$ down.

You chose B: Takes the middle $y$-value to be halfway between the two outer ones, though $x = 1$ is not halfway between $-2$ and $7$.

You chose C: Works back from $(7, -5)$ with the slope's sign dropped, falling a further $\tfrac{4}{3}$ per unit across the six units to $x = 1$.

Solution

Correct answer: C

You chose A: Applies the product test to a pair in which one of the two lines has no slope number to multiply.

You chose B: Reads "not slanted" as one shared direction, though one line is flat and the other stands upright.

You chose D: Requires the slope test to certify every right angle, so a pair the test cannot reach is ruled out rather than settled another way.

Solution

Correct answer: B

You chose A: Distributes the $-3$ over the parenthesis as $-3x + 24$, flipping the sign of the second term.

You chose C: Gathers the variable terms onto the side that leaves the leading coefficient negative, which the convention $A \ge 0$ rules out.

You chose D: Carries the $-20$ to the right-hand side without changing its sign, so the two constants add rather than partly cancel.

Solution

Correct answer: D

You chose A: Solves for $c$ as $-15$ by dropping the minus sign on the $y$-term, then divides that by $8$.

You chose B: Divides the coefficient by the constant rather than the constant by the coefficient.

You chose C: Uses the y-intercept's height $-5$ as the value of $c$, without solving $-3y = c$ for it first.

Solution

Correct answer: A

You chose B: Forgets that carrying $kx$ across the equals sign changes its sign, and so solves $\tfrac{k}{3} = -4$.

You chose C: Divides the required slope by $3$ rather than multiplying by it.

You chose D: Reads the coefficient of $x$ as the slope without solving for $y$ first, so the required slope is copied straight into $k$.

Solution

Correct answer: B

You chose A: Keeps the given line's own slope, which builds a line alongside it rather than one across it.

You chose C: Turns the fraction over but leaves the sign negative, so the product of the slopes comes out $+1$.

You chose D: Changes the sign of the slope without turning the fraction over, so the product of the slopes comes out $-\tfrac{25}{4}$.

Solution

Correct answer: C

You chose A: Reads the two numbers straight off the equation without reversing the signs that its subtractions impose.

You chose B: Pairs the two constants in the order they are printed, taking the number beside $y$ as the first coordinate.

You chose D: Assumes the shared point must be the y-intercept, reading the $3$ as the height at $x = 0$.