What is Coordinate geometry on the Geometry Regents?

Coordinate geometry is a Coordinate Geometry topic on the Geometry Regents. RegentsLab has 34 practice questions on it, each with a full worked solution.

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Coordinate geometry: Geometry Regents Practice

Practice Coordinate geometry for the Geometry Regents with real-style questions and full worked solutions. Miss one? Read the solution — the exam reuses the same ideas every year.

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Question 8 · 2 pts

What is the midpoint of the segment with endpoints $(2,\,-3)$ and $(8,\,5)$ ?

(A) $(6,\,8)$
(B) $(10,\,2)$
(C) $(5,\,1)$ ✓
(D) $(3,\,4)$

Worked solution

$\left(\frac{2+8}{2},\ \frac{-3+5}{2}\right)$ Midpoint = average of the coordinates.

$= (5,\,1)$

Answer: (5,\,1)

The midpoint averages the x-coordinates and the y-coordinates separately.

Question 33 · 2 pts

What are the coordinates of the midpoint of $\overline{AB}$ with endpoints $A(-3,5)$ and $B(7,-1)$?

(A) $(4,\ 4)$
(B) $(5,\ -3)$
(C) $(2,\ 2)$ ✓
(D) $(-5,\ 3)$

Worked solution

$M=\left(\dfrac{x_1+x_2}{2},\ \dfrac{y_1+y_2}{2}\right)$ Midpoint formula

$M=\left(\dfrac{-3+7}{2},\ \dfrac{5+(-1)}{2}\right)$

$M=\left(\dfrac{4}{2},\ \dfrac{4}{2}\right)=(2,\ 2)$

Answer: $(2,\ 2)$

The midpoint of a segment is found by averaging the $x$-coordinates and averaging the $y$-coordinates of the two endpoints.

Question 34 · 2 pts

What is the distance between the points $P(2,-3)$ and $Q(-2,0)$?

(A) $\sqrt{7}$
(B) $7$
(C) $25$
(D) $5$ ✓

Worked solution

$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$ Distance formula

$d=\sqrt{(2-(-2))^2+(-3-0)^2}$

$d=\sqrt{4^2+(-3)^2}=\sqrt{16+9}$

$d=\sqrt{25}=5$

Answer: $5$

The distance formula, $d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$, is the Pythagorean Theorem applied to the horizontal and vertical legs between two points.

Question 16 · 2 pts

The midpoint of $\overline{AB}$ is $M(1,-2)$. If the coordinates of $A$ are $(-3,4)$, what are the coordinates of $B$?

(A) $(-7,10)$
(B) $(-1,1)$
(C) $(5,-8)$ ✓
(D) $(4,-6)$

Worked solution

$M = \left(\frac{x_A + x_B}{2}, \frac{y_A + y_B}{2}\right)$ Midpoint formula.

$1 = \frac{-3 + x_B}{2} \Rightarrow x_B = 5$

$-2 = \frac{4 + y_B}{2} \Rightarrow y_B = -8$

$B = (5,-8)$

Answer: $(5,-8)$

Since $M$ is the midpoint, each coordinate of $M$ is the average of the corresponding coordinates of $A$ and $B$. Solving $\frac{-3+x}{2}=1$ gives $x=5$, and solving $\frac{4+y}{2}=-2$ gives $y=-8$. The point $(-1,1)$ is the midpoint of $A$ and $M$, a common error, and $(-7,10)$ results from extending in the wrong direction.

Question 9 · 2 pts

What is the midpoint of the segment joining $(2,3)$ and $(8,11)$?

(A) $(10,14)$
(B) $(6,8)$
(C) $(3,4)$
(D) $(5,7)$ ✓

Worked solution

$M=\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)$ Midpoint formula.

$=\left(\frac{2+8}{2},\frac{3+11}{2}\right)=(5,7)$

Answer: (5,7)

The midpoint is the average of the x-coordinates and the average of the y-coordinates.

Question 18 · 2 pts

A line has slope $\tfrac{2}{3}$. What is the slope of a line perpendicular to it?

(A) $\tfrac{3}{2}$
(B) $-\tfrac{2}{3}$
(C) $\tfrac{2}{3}$
(D) $-\tfrac{3}{2}$ ✓

Worked solution

$m_\perp=-\frac{1}{m}$ Perpendicular slopes are negative reciprocals.

$-\frac{1}{2/3}=-\frac{3}{2}$

Answer: -\tfrac{3}{2}

Perpendicular lines have slopes that are negative reciprocals: flip the fraction and change the sign.

Question 17 · 2 pts

What are the center and radius of the circle $(x-2)^2+(y+3)^2=16$?

(A) center $(2,-3)$, radius $4$ ✓
(B) center $(2,-3)$, radius $16$
(C) center $(-2,3)$, radius $16$
(D) center $(-2,3)$, radius $4$

Worked solution

$(x-h)^2+(y-k)^2=r^2$ Standard circle form.

$h=2,\ k=-3,\ r=\sqrt{16}=4$

Answer: (2,-3),\ r=4

In (x−h)²+(y−k)²=r², the center is (h,k), read the opposite signs, and the radius is the square root of the right side.

Question 34 · 4 pts · constructed response

Triangle $ABC$ has vertices $A(1,1)$, $B(5,1)$, and $C(5,4)$. Determine the length of $AC$.

Model answer: 5

Worked solution

$AC=\sqrt{(5-1)^2+(4-1)^2}$ Distance formula.

$=\sqrt{16+9}=\sqrt{25}=5$

Answer: AC=5

Apply the distance formula to A(1,1) and C(5,4): √(4² + 3²) = √25 = 5.

Question 35 · 2 pts

A line passes through the points $(1,2)$ and $(5,-6)$. What is the slope of a line perpendicular to this line?

(A) $-2$
(B) $-\dfrac{1}{2}$
(C) $\dfrac{1}{2}$ ✓
(D) $2$

Worked solution

$m=\dfrac{y_2-y_1}{x_2-x_1}=\dfrac{-6-2}{5-1}=\dfrac{-8}{4}=-2$

$m_\perp=-\dfrac{1}{m}=-\dfrac{1}{-2}=\dfrac{1}{2}$ Negative reciprocal

Answer: $\dfrac{1}{2}$

Perpendicular lines have slopes that are negative reciprocals, so you flip the fraction and change its sign.

Question 36 · 2 pts

Which equation represents the line that passes through the point $(4,-1)$ and is parallel to the line $y=2x+5$?

(A) $y=2x+7$
(B) $y=2x-1$
(C) $y=-\dfrac{1}{2}x+1$
(D) $y=2x-9$ ✓

Worked solution

$\text{Parallel lines share a slope, so } m=2$

$y-y_1=m(x-x_1)\ \Rightarrow\ y-(-1)=2(x-4)$

$y+1=2x-8$

$y=2x-9$

Answer: $y=2x-9$

Parallel lines have equal slopes; substitute the shared slope and the given point into point-slope form, then solve for $y$.

Question 39 · 2 pts

The point $M(3,-1)$ is the midpoint of $\overline{AB}$. If $A$ has coordinates $(-2,4)$, what are the coordinates of $B$?

(A) $(-7,\ 9)$
(B) $(5,\ -5)$
(C) $(0.5,\ 1.5)$
(D) $(8,\ -6)$ ✓

Worked solution

$M=\dfrac{A+B}{2}\ \Rightarrow\ B=2M-A$ Solve the midpoint formula for the missing endpoint

$B=(2(3)-(-2),\ 2(-1)-4)$

$B=(6+2,\ -2-4)$

$B=(8,\ -6)$

Answer: $(8,\ -6)$

When a midpoint and one endpoint are known, solve $M=\frac{A+B}{2}$ for the missing endpoint to get $B=2M-A$.

Question 15 · 2 pts

Which equation represents the line that is perpendicular to $y = \frac{2}{3}x - 5$ and passes through the point $(4, -1)$?

(A) $y + 1 = \frac{2}{3}(x - 4)$
(B) $y + 1 = \frac{3}{2}(x - 4)$
(C) $y - 1 = -\frac{3}{2}(x + 4)$
(D) $y + 1 = -\frac{3}{2}(x - 4)$ ✓

Worked solution

$m_{\text{given}} = \frac{2}{3}$ Slope of the given line.

$m_{\perp} = -\frac{3}{2}$ Perpendicular slopes are negative reciprocals.

$y - (-1) = -\frac{3}{2}(x - 4)$ Point-slope form through (4, -1).

$y + 1 = -\frac{3}{2}(x - 4)$

Answer: $y + 1 = -\frac{3}{2}(x - 4)$

A perpendicular line has slope equal to the negative reciprocal of 2/3, which is -3/2. Substituting the point (4, -1) into point-slope form gives y + 1 = -3/2(x - 4). The distractors use the same slope, the reciprocal without the sign change, or the wrong signs on the point.

Question 31 · 4 pts · constructed response

The endpoints of directed line segment $AB$ are $A(-4, 2)$ and $B(6, 7)$. Determine the coordinates of point $P$ on $\overline{AB}$ such that $P$ divides the segment in the ratio $AP:PB = 2:3$.

Model answer: $P(0, 4)$

Worked solution

$P = A + \frac{2}{5}(B - A)$ A ratio of 2:3 means P is 2/5 of the way from A to B.

$x_P = -4 + \frac{2}{5}(6 - (-4)) = -4 + \frac{2}{5}(10) = -4 + 4 = 0$

$y_P = 2 + \frac{2}{5}(7 - 2) = 2 + \frac{2}{5}(5) = 2 + 2 = 4$

$P(0, 4)$

Answer: $P(0, 4)$

A point dividing a directed segment in the ratio $2:3$ lies $\frac{2}{2+3} = \frac{2}{5}$ of the way from $A$ to $B$. Adding $\frac{2}{5}$ of the horizontal and vertical changes to the coordinates of $A$ gives $P(0,4)$.

Question 4 · 2 pts

What is an equation of the line that passes through the point $(6, -1)$ and is perpendicular to the line whose equation is $y = \frac{2}{3}x - 4$?

(A) $y = -\frac{3}{2}x + 8$ ✓
(B) $y = \frac{2}{3}x - 5$
(C) $y = \frac{3}{2}x - 10$
(D) $y = -\frac{3}{2}x - 10$

Worked solution

$m_{\perp} = -\frac{3}{2}$ Perpendicular slopes are negative reciprocals of each other.

$y - (-1) = -\frac{3}{2}(x - 6)$ Use point-slope form with the point (6, -1).

$y + 1 = -\frac{3}{2}x + 9$

$y = -\frac{3}{2}x + 8$

Answer: $y = -\frac{3}{2}x + 8$

A perpendicular line has slope equal to the negative reciprocal of $\frac{2}{3}$, which is $-\frac{3}{2}$. Substituting the point $(6,-1)$ into point-slope form and simplifying gives $y = -\frac{3}{2}x + 8$. Check: $-\frac{3}{2}(6) + 8 = -9 + 8 = -1$. Using the same slope or forgetting the negative sign produces the distractors.

Question 20 · 2 pts

Which equation represents the line that passes through the point $(3, -2)$ and is perpendicular to the line whose equation is $y = \frac{2}{3}x + 5$?

(A) $y + 2 = \frac{3}{2}(x - 3)$
(B) $y - 2 = -\frac{3}{2}(x + 3)$
(C) $y + 2 = \frac{2}{3}(x - 3)$
(D) $y + 2 = -\frac{3}{2}(x - 3)$ ✓

Worked solution

$m_{\perp} = -\frac{3}{2}$ Perpendicular slopes are negative reciprocals; the negative reciprocal of $\frac{2}{3}$ is $-\frac{3}{2}$.

$y - y_1 = m(x - x_1)$ Use point-slope form with $(3, -2)$.

$y - (-2) = -\frac{3}{2}(x - 3)$

$y + 2 = -\frac{3}{2}(x - 3)$

Answer: $y + 2 = -\frac{3}{2}(x - 3)$

A perpendicular line has a slope that is the negative reciprocal of the original slope, so the new slope is $-\frac{3}{2}$. Substituting the point $(3, -2)$ into point-slope form gives $y + 2 = -\frac{3}{2}(x - 3)$. Watch the signs: $y - (-2)$ becomes $y + 2$.

Question 28 · 4 pts · constructed response

The endpoints of directed line segment $\overline{AB}$ are $A(-4,1)$ and $B(6,6)$. Determine and state the coordinates of point $P$ on $\overline{AB}$ such that $AP:PB = 2:3$.

Model answer: $P(0,3)$

Worked solution

$P = A + \frac{2}{5}(B - A)$ The ratio $2:3$ means $P$ is $\frac{2}{5}$ of the way from $A$ to $B$.

$x_P = -4 + \frac{2}{5}(6-(-4)) = -4 + \frac{2}{5}(10) = 0$

$y_P = 1 + \frac{2}{5}(6-1) = 1 + 2 = 3$

$P(0,3)$

Answer: $P(0,3)$

A ratio of $AP:PB = 2:3$ places $P$ two fifths of the way along the directed segment from $A$ to $B$. Adding $\frac{2}{5}$ of the change in each coordinate to the coordinates of $A$ gives $P(0,3)$.

Question 33 · 4 pts · constructed response

Quadrilateral $ABCD$ has vertices $A(1,2)$, $B(4,6)$, $C(9,6)$, and $D(6,2)$. Prove that $ABCD$ is a rhombus. Then prove that $ABCD$ is not a square. [The use of the set of axes is optional; all work must be shown using coordinate methods.]

Model answer: All four sides have length $5$ ($AB = BC = CD = DA = 5$ by the distance formula), so $ABCD$ is a rhombus. The slope of $\overline{AB}$ is $\frac{4}{3}$ and the slope of $\overline{BC}$ is $0$; since $\frac{4}{3} \cdot 0 \neq -1$, consecutive sides are not perpendicular, so $ABCD$ has no right angle and is not a square.

Worked solution

$AB = \sqrt{(4-1)^2 + (6-2)^2} = \sqrt{9+16} = 5$ Use the distance formula on each side.

$BC = \sqrt{(9-4)^2 + (6-6)^2} = \sqrt{25} = 5$

$CD = \sqrt{(6-9)^2 + (2-6)^2} = \sqrt{9+16} = 5$

$DA = \sqrt{(1-6)^2 + (2-2)^2} = \sqrt{25} = 5$

$AB = BC = CD = DA = 5, \text{ so } ABCD \text{ is a rhombus}$ A quadrilateral with four congruent sides is a rhombus.

$\text{slope of } \overline{AB} = \frac{6-2}{4-1} = \frac{4}{3}, \quad \text{slope of } \overline{BC} = \frac{6-6}{9-4} = 0$ Check whether consecutive sides are perpendicular.

$\frac{4}{3} \cdot 0 \neq -1, \text{ so } \overline{AB} \not\perp \overline{BC}, \text{ and } ABCD \text{ is not a square}$ A square requires a right angle; a horizontal side is only perpendicular to a vertical side.

Answer: All four sides have length $5$ ($AB = BC = CD = DA = 5$ by the distance formula), so $ABCD$ is a rhombus. The slope of $\overline{AB}$ is $\frac{4}{3}$ and the slope of $\overline{BC}$ is $0$; since $\frac{4}{3} \cdot 0 \neq -1$, consecutive sides are not perpendicular, so $ABCD$ has no right angle and is not a square.

The distance formula shows all four sides are congruent, which is exactly the definition of a rhombus. To rule out a square, it is enough to show one pair of consecutive sides is not perpendicular; the slopes $\frac{4}{3}$ and $0$ do not multiply to $-1$, so the angle at $B$ is not a right angle.

Question 11 · 2 pts

Which equation represents the line that is perpendicular to $y = \frac{2}{3}x + 5$ and passes through the point $(6, -1)$?

(A) $y = -\frac{2}{3}x + 3$
(B) $y = -\frac{3}{2}x + 8$ ✓
(C) $y = \frac{2}{3}x - 5$
(D) $y = \frac{3}{2}x - 10$

Worked solution

$\text{Perpendicular slopes are negative reciprocals: } m = -\frac{3}{2}$

$y - (-1) = -\frac{3}{2}(x - 6)$ Point-slope form through (6, -1).

$y + 1 = -\frac{3}{2}x + 9$

$y = -\frac{3}{2}x + 8$ Check: at x = 6, y = -9 + 8 = -1.

Answer: $y = -\frac{3}{2}x + 8$

A perpendicular line has slope equal to the negative reciprocal of 2/3, which is -3/2. Using point-slope form through (6, -1) and simplifying gives y = -3/2 x + 8. The distractors use the same slope (parallel), the reciprocal without the sign change, or the opposite slope without the reciprocal.

Question 24 · 2 pts

The endpoints of directed line segment $AB$ are $A(-4, 1)$ and $B(6, 6)$. What are the coordinates of point $P$ on $\overline{AB}$ such that $AP:PB = 2:3$?

(A) $(0, 3)$ ✓
(B) $(-2, 2)$
(C) $(1, 3.5)$
(D) $(2, 4)$

Worked solution

$AP:PB = 2:3 \text{ means } P \text{ is } \tfrac{2}{5} \text{ of the way from } A \text{ to } B$ The whole segment is divided into 2 + 3 = 5 equal parts.

$P = \left(x_A + \tfrac{2}{5}(x_B - x_A),\; y_A + \tfrac{2}{5}(y_B - y_A)\right)$

$x_P = -4 + \tfrac{2}{5}(6 - (-4)) = -4 + \tfrac{2}{5}(10) = -4 + 4 = 0$

$y_P = 1 + \tfrac{2}{5}(6 - 1) = 1 + \tfrac{2}{5}(5) = 1 + 2 = 3$

$P = (0, 3)$

Answer: $(0, 3)$

A ratio $AP:PB = 2:3$ places $P$ at $\frac{2}{5}$ of the way along the directed segment from $A$ to $B$. Adding $\frac{2}{5}$ of the change in each coordinate to $A$ gives $(0, 3)$. Choosing $(2, 4)$ comes from reversing the ratio, $(1, 3.5)$ is the midpoint, and $(-2, 2)$ comes from using $\frac{1}{5}$ instead of $\frac{2}{5}$.

Question 31 · 4 pts · constructed response

The endpoints of $\overline{AB}$ are $A(-2,3)$ and $B(4,-1)$. Determine and state an equation of the perpendicular bisector of $\overline{AB}$.

Model answer: $y - 1 = \frac{3}{2}(x - 1)$ (equivalently $y = \frac{3}{2}x - \frac{1}{2}$)

Worked solution

$M = \left(\frac{-2+4}{2}, \frac{3+(-1)}{2}\right) = (1, 1)$ The perpendicular bisector passes through the midpoint.

$\text{slope of } \overline{AB} = \frac{-1-3}{4-(-2)} = \frac{-4}{6} = -\frac{2}{3}$

$\text{perpendicular slope} = \frac{3}{2}$ Perpendicular slopes are negative reciprocals.

$y - 1 = \frac{3}{2}(x - 1)$

Answer: $y - 1 = \frac{3}{2}(x - 1)$ (equivalently $y = \frac{3}{2}x - \frac{1}{2}$)

The perpendicular bisector of a segment passes through the segment's midpoint and is perpendicular to it. The midpoint of $\overline{AB}$ is $(1,1)$ and the slope of $\overline{AB}$ is $-\frac{2}{3}$, so the bisector has slope $\frac{3}{2}$, giving $y - 1 = \frac{3}{2}(x-1)$.

Question 5 · 2 pts

What is the slope of a line perpendicular to the line whose equation is $3x + 4y = 8$?

(A) $-\frac{4}{3}$
(B) $\frac{3}{4}$
(C) $-\frac{3}{4}$
(D) $\frac{4}{3}$ ✓

Worked solution

$4y = -3x + 8 \;\Rightarrow\; y = -\tfrac{3}{4}x + 2$ Solve for $y$ to find the slope of the given line.

$m = -\tfrac{3}{4}$

$m_{\perp} = \tfrac{4}{3}$ Perpendicular slopes are negative reciprocals.

Answer: $\frac{4}{3}$

Rewriting 3x + 4y = 8 in slope-intercept form shows the slope is -3/4. A perpendicular line has the negative reciprocal slope, which is 4/3. Choosing -3/4 gives a parallel line, and 3/4 or -4/3 come from forgetting to flip or to change the sign.

Question 20 · 2 pts

What is the equation of a circle with center $(3,\,-2)$ and radius $5$ ?

(A) $(x+3)^2 + (y-2)^2 = 25$
(B) $(x-3)^2 + (y+2)^2 = 25$ ✓
(C) $(x-3)^2 + (y-2)^2 = 25$
(D) $(x-3)^2 + (y+2)^2 = 5$

Worked solution

$(x-h)^2 + (y-k)^2 = r^2$ Standard circle equation, center (h, k).

$(x-3)^2 + (y+2)^2 = 25$

Answer: (x-3)^2+(y+2)^2=25

The center's coordinates enter with opposite signs, and the right side is the radius squared.

Question 37 · 2 pts

Directed line segment $AB$ has endpoints $A(-2,1)$ and $B(4,10)$. What are the coordinates of point $P$ that partitions $\overline{AB}$ so that $AP:PB=1:2$?

(A) $(1,\ 5.5)$
(B) $(0,\ 4)$ ✓
(C) $(2,\ 7)$
(D) $(-4,\ -2)$

Worked solution

$P=A+\dfrac{1}{1+2}(B-A)$ AP:PB = 1:2, so P is 1/3 of the way from A to B

$B-A=(4-(-2),\ 10-1)=(6,\ 9)$

$P=(-2,1)+\dfrac{1}{3}(6,9)=(-2+2,\ 1+3)$

$P=(0,\ 4)$

Answer: $(0,\ 4)$

A point that partitions a directed segment in ratio $a:b$ lies $\frac{a}{a+b}$ of the way from the first endpoint toward the second.

Question 38 · 2 pts

What is an equation of the perpendicular bisector of the segment with endpoints $A(1,1)$ and $B(5,3)$?

(A) $y=-2x+8$ ✓
(B) $y=-2x+6$
(C) $y=2x-4$
(D) $y=\dfrac{1}{2}x+\dfrac{1}{2}$

Worked solution

$\text{Midpoint}=\left(\dfrac{1+5}{2},\ \dfrac{1+3}{2}\right)=(3,\ 2)$

$m_{AB}=\dfrac{3-1}{5-1}=\dfrac{2}{4}=\dfrac{1}{2}$

$m_\perp=-2$ Negative reciprocal of 1/2

$y-2=-2(x-3)\ \Rightarrow\ y=-2x+6+2$

$y=-2x+8$

Answer: $y=-2x+8$

The perpendicular bisector passes through the midpoint of a segment and has a slope equal to the negative reciprocal of the segment's slope.

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