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GRE Subject Test: Math : Distance & Midpoint Formulas

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Example Questions

Example Question #21 : Distance & Midpoint Formulas

Given two points, (1,7) and (6,11) , find the midpoint.

Possible Answers:

(9,-3.5)

(-3.5,9)

(9,3.5)

(3.5,9)

Correct answer:

(3.5,9)

Explanation:

Step 1: Define midpoint. The midpoint is a point located between two given points.. If I draw a line through these points, I get a straight line

Step 2: The midpoint formula is: $\bigg(\frac {x_1+x_2}{2},\frac {y_1+y_2}{2}\bigg)$

Step 3. Plug in the values:

$\bigg(\frac {1+6}{2},\frac {7+11}{2}\bigg)$

Step 4: Simplify each fraction in Step 3:

$\bigg(\frac {7}{2},\frac {18}{2}\bigg)$

Step 5: Convert each fraction to a decimal from step 4:

(3.5,9)

The midpoint is (3.5,9)

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Example Question #32 : Calculus

Find the length of the line that connects the points: (4,7) and (8,15) .

Possible Answers:

$-4\sqrt{10}$

$4\sqrt{157}$

$4\sqrt{5}$

$\sqrt{586}$

Correct answer:

$4\sqrt{5}$

Explanation:

Step 1: Recall the distance formula:

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

Step 2: Find x_1, x_2, y_1, y_2 ...

x_1=4, x_2=8, y_1=7, y_2=15

Step 3: Substitute the values into the equation:

$d=\sqrt{(8-4)^2+(15-7)^2}$
Reduce the parentheses:

$d=\sqrt{4^2+8^2}$
Evaluate the exponents and add:

$d=\sqrt{16+64}=\sqrt{80}$

Step 4: Reduce $\sqrt{80}$ into lowest terms...

$\sqrt{80}=\sqrt{4}\cdot \sqrt{4}\cdot \sqrt 5$

Using rule of square roots, multiplying two roots with the same value on the inside just gives me the inside value..

$\sqrt{80}=[\sqrt{4}\cdot \sqrt{4}]\cdot \sqrt{5}$
$\sqrt{80}=4\sqrt{5}$

The length of the line that connects both points is $4\sqrt{5}$ .

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Example Question #21 : Distance & Midpoint Formulas

What is the distance between (4,6) and (11,7)

Possible Answers:

$5\sqrt{3}$

$5\sqrt{2}$

$2\sqrt{5}$

Correct answer:

$5\sqrt{2}$

Explanation:

Step 1: Plug in values into the distance formula:

$\sqrt{({11-4})^2+(7-6)^2}$

Step 2: Evaluate the inside...

$\sqrt{7^2+1^2}=\sqrt{50}$

Step 3: Simplify...

$\sqrt{50}=\sqrt{25}\sqrt{2}=5\sqrt2$

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Example Question #32 : Functions And Graphs

Find the midpoint between (5,-2) and (9,-4)

Possible Answers:

(7,-3)

(-2, -1)

(3,3/2)

(3,-6)

(3,3)

Correct answer:

(7,-3)

Explanation:

To find the midpoint you must use the equation

$\left (\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2} \right )$ Insert numbers

$\left (\frac{5+9}{2}, \frac{-2+-4}{2} \right )$

(7,-3)

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Example Question #22 : Distance & Midpoint Formulas

Find the distance between the two points

(3,-4), (2,5)

Possible Answers:

$\sqrt{82}$

$\sqrt{58}$

$\sqrt{93}$

Correct answer:

$\sqrt{82}$

Explanation:

You must use the distance formula

$\sqrt{({x}_{2}+{x}_{1})^{2}+({y}_{2}+{y}_{1})^{2}}$ Fill in with the points

$\sqrt{({2}+{3})^{2}+({5}+{-4})^{2}}}$ Solve

$\sqrt{({-1})^{2}+({9})^{2}}$

$\sqrt{{1}+{81}}$

$\sqrt{82}$

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GRE Subject Test: Math : Distance & Midpoint Formulas

Study concepts, example questions & explanations for GRE Subject Test: Math

All GRE Subject Test: Math Resources

Example Questions

Example Question #21 : Distance & Midpoint Formulas

Example Question #32 : Calculus

Example Question #21 : Distance & Midpoint Formulas

Example Question #32 : Functions And Graphs

Example Question #22 : Distance & Midpoint Formulas

All GRE Subject Test: Math Resources

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