GRE Quantitative Flashcards: Coordinate Geometry Distance

What this deck covers

This deck focuses on Coordinate Geometry Distance, giving you a quick way to review the definitions, rules, and examples that matter most for GRE Quantitative.

How to use these flashcards

Work through these flashcards in short sessions. Try to answer each prompt before flipping the card, then revisit any cards you miss until the explanation feels automatic.

Flashcard 1: What is the distance between $(-1,-2)$ and $(2,2)$?

Answer: $5$. Distance formula results in $sqrt{(2-(-1))^2+(2-(-2))^2}=sqrt{9+16}=5$.

Flashcard 2: What is the midpoint of the segment joining $(-3,5)$ and $(1,-1)$?

Answer: $\left(-1,2\right)$. Midpoint formula: left(rac{-3+1}{2},rac{5+(-1)}{2} ight)=(-1,2).

Flashcard 3: What is the distance between $(1,2)$ and $(4,6)$?

Answer: $5$. Distance formula gives $sqrt{(4-1)^2+(6-2)^2}=sqrt{9+16}=5$.

Flashcard 4: State the formula for the midpoint of the segment joining $(x_1,y_1)$ and $(x_2,y_2)$.

Answer: $\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)$. Averages the x-coordinates and y-coordinates separately to find the central point of the line segment.

Flashcard 5: What is the equation of the circle with center $(0,0)$ and radius $r$?

Answer: $x^2+y^2=r^2$. Special case of the circle equation when the center is at the origin, using distance from $(0,0)$.

Flashcard 6: What is the distance from $(4,-3)$ to the origin $(0,0)$?

Answer: $5$. Distance to origin: $sqrt{(4-0)^2+(-3-0)^2}=sqrt{16+9}=5$.

Flashcard 7: Identify the condition for points $(x_1,y_1)$ and $(x_2,y_2)$ to be exactly $r$ units apart.

Answer: $(x_2-x_1)^2+(y_2-y_1)^2=r^2$. Sets the squared distance equal to $r^2$ to define points exactly $r$ units apart without needing the square root.

Flashcard 8: What is the distance between $(5,-1)$ and $(5,6)$?

Answer: $7$. Vertical line segment with $Delta x=0$ and $Delta y=7$, so distance equals $|6-(-1)|$.

Flashcard 9: What is the distance between $(2,3)$ and $(8,3)$?

Answer: $6$. Horizontal line segment with $Delta x=6$ and $Delta y=0$, so distance equals $|8-2|$.

Flashcard 10: What is the equation of the circle with center $(h,k)$ and radius $r$?

Answer: $(x-h)^2+(y-k)^2=r^2$. Represents all points $(x,y)$ at distance $r$ from center $(h,k)$ using the distance formula squared.

Flashcard 11: State the formula for the distance from the origin $(0,0)$ to a point $(x,y)$.

Answer: $\sqrt{x^2+y^2}$. Applies the distance formula with $(x_1,y_1)=(0,0)$, yielding the magnitude of the position vector from the origin.

Flashcard 12: What is the distance from $(x_1,y_1)$ to $(x_1,y_2)$ (same $x$-coordinate)?

Answer: $|y_2-y_1|$. With identical x-coordinates, the distance simplifies to the absolute difference in y-coordinates, as the horizontal change is zero.

Flashcard 13: What is the distance from point $(x_0,y_0)$ to the horizontal line $y=c$?

Answer: $|y_0-c|$. Vertical distance between the point's y-coordinate and the line $y=c$ uses the absolute difference.

Flashcard 14: State the distance formula between two points $(x_1,y_1)$ and $(x_2,y_2)$ in the coordinate plane.

Answer: $d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$. Derived from the Pythagorean theorem, calculating the hypotenuse of the right triangle formed by the horizontal and vertical differences in coordinates.

Flashcard 15: What is the midpoint of the segment joining $(2,7)$ and $(6,1)$?

Answer: $\left(4,4\right)$. Midpoint formula: left(rac{2+6}{2},rac{7+1}{2} ight)=(4,4).

Flashcard 16: What is the squared distance between $(x_1,y_1)$ and $(x_2,y_2)$ (no square root)?

Answer: $(x_2-x_1)^2+(y_2-y_1)^2$. Squares the differences in coordinates to compute distance squared, avoiding the square root for comparisons or equations.

Flashcard 17: What is the distance between $(-2,1)$ and $(1,5)$?

Answer: $5$. Distance formula yields $sqrt{(1-(-2))^2+(5-1)^2}=sqrt{9+16}=5$.

Flashcard 18: What is the distance from $(x_1,y_1)$ to $(x_2,y_1)$ (same $y$-coordinate)?

Answer: $|x_2-x_1|$. With identical y-coordinates, the distance simplifies to the absolute difference in x-coordinates, as the vertical change is zero.

Flashcard 19: What is the distance from $(x,y)$ to the $x$-axis?

Answer: $|y|$. Vertical distance from any point on the line $y=y$ to $y=0$ is the absolute value of the y-coordinate.

Flashcard 20: What is the length of the diagonal of a rectangle with corners $(0,0)$ and $(6,8)$?

Answer: $10$. Diagonal length is the distance between opposite corners: $sqrt{(6-0)^2+(8-0)^2}=sqrt{36+64}=10$.

Flashcard 21: What is the distance from $(x,y)$ to the $y$-axis?

Answer: $|x|$. Horizontal distance from any point on the line $x=x$ to $x=0$ is the absolute value of the x-coordinate.

Flashcard 22: What is the distance between $(0,0)$ and $(3,4)$?

Answer: $5$. Applies the distance formula: $sqrt{(3-0)^2+(4-0)^2}=sqrt{9+16}=sqrt{25}=5$.

Flashcard 23: What is the distance from point $(x_0,y_0)$ to the vertical line $x=c$?

Answer: $|x_0-c|$. Horizontal distance between the point's x-coordinate and the line $x=c$ uses the absolute difference.