GMAT Math : Geometry

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Example Questions

Example Question #5 : Midpoint Formula

If the midpoint of $\overline{VU}$ is $\left ( 6,7\right )$ and $\textup{Point }V$ is at $\left (2,9\right )$ , what are the coordinates of $\textup{Point }U$ ?

Possible Answers:

$\left ( 14,23\right )$

$\left ( 23,14\right )$

$\left ( 10,5\right )$

$\left ( 5,10\right )$

$\left ( -5,10\right )$

Correct answer:

$\left ( 10,5\right )$

Explanation:

Midpoint formula is as follows:

$\small \small m(x,y)=(\frac{x+x'}{2},\frac{y+y'}{2})$

In this case, we have x,y and the value of the midpoint. We need to findx' and y'

V is at (2,9) and the midpoint is at (6,7)

$\small 6=\frac{2+x'}{2}$ and $\small 7=\frac{9+y'}{2}$

$\small x'=6*2-2=10$ $\small y'=7*2-9=14-9=5$

So we have (10,5) as point U

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Example Question #11 : Coordinate Geometry

A line segment has its midpoint at (-1,3) and an endpoint at (3,8) . What are the coordinates of the other endpoint?

Possible Answers:

(-3,-8)

(-2,-5)

(-5,-2)

(-3,-1)

(-2,1)

Correct answer:

(-5,-2)

Explanation:

Because we are given the midpoint and one of the endpoints, we know the x coordinate of the other endpoint will be the same distance away from the midpoint in the x direction, and the y coordinate of the other endpoint will be the same distance away from the midpoint in the y direction. Given two endpoints of the form:

(x_1,y_1),(x_2,y_2)

The midpoint of these two endpoints has the coordinates:

$(x_m,y_m)=(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})$

Plugging in values for the given midpoint and one of the endpoints, which we can see is (x_2,y_2) because it lies to the right of the midpoint, we can solve for the other endpoint as follows:

$(x_m,y_m)=(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})$

$(-1,3)=(\frac{x_1+3}{2},\frac{y_1+8}{2})$

$\frac{x_1+3}{2}=-1\rightarrow x_1+3=-2\rightarrow x_1=-5$

$\frac{y_1+8}{2}=3\rightarrow y_1+8=6\rightarrow y_1=-2$

So the other endpoint has the coordinates (-5,-2)

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Example Question #2 : Calculating The Endpoints Of A Line Segment

Consider segment with endpoint at (8,9) . If the midpoint of can be found at (-15, -67) , what are the coordinates of point ?

Possible Answers:

(38,143)

(-38,-143)

(-46,-150)

(143,38)

(-143,-38)

Correct answer:

(-38,-143)

Explanation:

Recall midpoint formula:

$(x',y')=\left(\frac{x^1+x^2}{2}, \frac{y^1+y^2}{2}\right)$

In this case we have (x'y') and one of our other (x,y) points.

Plug and chug:

$(-15,-67)=\left(\frac{8+x}{2}, \frac{9+y}{2}\right)$

If you make this into two equations and solve you get the following.

$\small -15=\frac{8+x}{2}, x=-38$

$\small -67=\frac{9+y}{2}, y=-143$

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Example Question #611 : Geometry

A line segement on the coordinate plane has endpoints (A,B+3) and (A + 4, B) . Which of the following expressions is equal to the length of the segment?

Possible Answers:

$\sqrt{(2A+4)^{2} + (2B+3)^{2}}$

$\sqrt{(2A-4)^{2} + (2B-3)^{2}}$

$\sqrt{(A-4)^{2} + (B-3)^{2}}$

$\sqrt{7}$

Correct answer:

Explanation:

Apply the distance formula, setting

$x_{ 1} = A + 4, x_{2} = A ,y_{ 1} = B, y_{2} = B+3$ :

$d = \sqrt{(x_{2}- x_{1})^{2}+(y_{2}-y_{1})^{2}}$

$d = \sqrt{[A-(A+4)]^{2}+[(B+3)-B)]^{2}}$

$d = \sqrt{(-4)^{2}+3^{2}} = \sqrt{16+9} = \sqrt{25} = 5$

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Example Question #612 : Geometry

What is distance between (6,7) and (1,5) ?

Possible Answers:

$\sqrt{20}$

$\sqrt{29}$

$\sqrt{7}$

$\sqrt{12}$

Correct answer:

$\sqrt{29}$

Explanation:

$distance= \sqrt{(x_{2}-x_{1})^2+y_{2}-y_{1})^2}$

$= \sqrt{(6-1)^2+(7-5)^2}$

$= \sqrt{(5)^2+(2)^2}$

$= \sqrt{(25+4)}$

$\sqrt{29}$

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Example Question #613 : Geometry

What is the distance between the points (2, 5) and (7, 17) ?

Possible Answers:

$2\sqrt{5}$

Correct answer:

Explanation:

Let's plug our coordinates into the distance formula.

$\sqrt{(2-7)^{2}+(5-17)^{2}}= \sqrt{(-5)^{2}+(-12)^{2}} = \sqrt{25+144}= \sqrt{169} = 13$

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Example Question #614 : Geometry

What is the distance between the points (1,4) and (5,2) ?

Possible Answers:

$2\sqrt{5}$

$3\sqrt{3}$

Correct answer:

$2\sqrt{5}$

Explanation:

We need to use the distance formula to calculate the distance between these two points.

$\sqrt{(1-5)^{2}+(4-2)^{2}} = \sqrt{(-4)^{2}+(2)^{2}} =\sqrt{20}=\sqrt{4}\sqrt{5}=2\sqrt{5}$

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Example Question #15 : Lines

Consider segment $\overline{JK}$ which passes through the points $\left ( 4,5\right )$ and $\left ( 144,75\right )$ .

Find the length of segment $\overline{JK}$ .

Possible Answers:

$169.4\ \textup{units}$

$134.6\ \textup{units}$

$176.5\ \textup{units}$

$168.2\ \textup{units}$

$156.5\ \textup{units}$

Correct answer:

$156.5\ \textup{units}$

Explanation:

This question requires careful application of distance formula, which is really a modified form of Pythagorean theorem.

$\small d=\sqrt{(x-x')^2+(y-y')^2}$

Plug in everthing and solve:

$\small \small d=\sqrt{(144-4)^2+(75-7)^2}=\sqrt{24500}\approx156.5$

So our answer is 156.6

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Example Question #16 : Lines

What is the length of a line segment that starts at the point (1,-3) and ends at the point (4,1) ?

Possible Answers:

Correct answer:

Explanation:

Using the distance formula for the length of a line between two points, we can plug in the given values and determine the length of the line segment by calculating the distance between the two points:

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

$d=\sqrt{(4-1)^2+(1-(-3))^2}$

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

$d=\sqrt{25}$

d=5

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Example Question #1 : Tangent Lines

Determine the equation of the line tangent to the curve y=x^2 at the point (-1,1) ?

Possible Answers:

y=-x+2

y=-2x-1

y=x-2

y=-2x+1

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

Correct answer:

y=-2x-1

Explanation:

To find the equation of a line tangent to a curve at a certain point, we first need to find the slope of the curve at that point. To find the slope of a function at any point, we need its derivative:

$y=x^2\rightarrow y'=2x$

Now we can plug in the x value of the given point to find the slope of the function, and therefore the slope of tangent line, at that point:

y'=2x

$y'(-1)=2(-1)=-2\rightarrow m=-2$

Now that we have our slope, we can simply plug this value in with the given point to solve for the y intercept of the tangent line:

y=mx+b

1=(-2)(-1)+b

b=-1

We have calculated the slope of the tangent line and its y intercept, so the equation for the line tangent to the curve y=x^2 at the point (-1,1) in standard form is:

y=-2x-1

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GMAT Math : Geometry

Study concepts, example questions & explanations for GMAT Math

All GMAT Math Resources

Example Questions

Example Question #5 : Midpoint Formula

Example Question #11 : Coordinate Geometry

Example Question #2 : Calculating The Endpoints Of A Line Segment

Example Question #611 : Geometry

Example Question #612 : Geometry

Example Question #613 : Geometry

Example Question #614 : Geometry

Example Question #15 : Lines

Example Question #16 : Lines

Example Question #1 : Tangent Lines

All GMAT Math Resources

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