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GRE Subject Test: Math : Distance & Midpoint Formulas

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

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Example Questions

Example Question #11 : Coordinate Geometry

Find the distance between $y = \frac{12}{5} x - 3$ and the point (3, -4)

Possible Answers:

$\frac{48}{5}$

$\frac{48}{13}$

$\frac{41}{5}$

$\frac{71}{13}$

$\frac{41}{13}$

Correct answer:

$\frac{41}{13}$

Explanation:

To find the distance, use the formula $distance = \frac{|ax_{0}+by_{0}+c |}{\sqrt{a^2 + b^2 }}$ where the point is (x_0, y_0) and the line is ax + by + c = 0

First, we'll re-write the equation $y = \frac{12}{5} x-3$ in this form to identify , , and :

$y = \frac{12}{5} x-3$ subtract $\frac{12}{5}x$ from and add to both sides

$y -\frac{12}{5}x +3 = 0$ multiply both sides by

5y -12x +15 = 0 Now we see that a = -12, b = 5, c = 15 . Plugging these plus (x_0, y_0) = (3,-4) into the formula, we get:

$distance = \frac{|-12(3)+5(-4)+15 |}{\sqrt{(5)^2 +( 12)^2 }} = \frac{|-36 + -20+15|}{\sqrt{25+144} } = \frac{41}{\sqrt{169}} = \frac{41}{13}$

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Example Question #1 : Find The Distance Between Two Parallel Lines

Find the distance between the two lines.

$\\ y_1=2.65x-13 \\ y_2=2.65x+4$

Possible Answers:

d=2.65

d=-17

d=9

d=6

Correct answer:

d=6

Explanation:

Since the slope of the two lines are equivalent, we know that the lines are parallel. Therefore, they are separated by a constant distance. We can then find the distance between the two lines by using the formula for the distance from a point to a nonvertical line:

$d=\frac{|Ax_1 +By_1+C|}{\sqrt{A^{2}+B^{2}}}$

First, we need to take one of the line and convert it to standard form.

y_1=2.65x-13

2.65x_1-y_1-13=0 where $A=2.65,B=-1, \textup{and }C=-13$

Now we can substitute A, B, and C into our distance equation along with a point, $\left ( x_1,y_1\right )$ , from the other line. We can pick any point we want, as long as it is on line y_2 . Just plug in a number for x, and solve for y. I will use the y-intercept, where x = 0, because it is easy to calculate:

y=2.65(0))+4

y=0+4

y=4

Now we have a point, (0,4) , that is on the line y_2 . So let's plug our values for $x_1,y_1,A,B,\textup{and }C}$ :

$d=\frac{|(2.65)\cdot (0) +(-1)\cdot (4)+(-13)|}{\sqrt{(2.65)^{2}+(-1)^{2}}}=\frac{|0-4-13|}{\sqrt{(7.0225+1)}}=\frac{|-17|}{\sqrt{8.0225}}$

$=\frac{17}{2.83}=6.002\approx6$

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Example Question #1 : Find The Distance Between Two Parallel Lines

Find the distance between y = 2x -3 and y = 2x+1

Possible Answers:

$\frac{6}{\sqrt5}$

$\frac{4}{\sqrt5}$

Correct answer:

$\frac{4}{\sqrt5}$

Explanation:

To find the distance, choose any point on one of the lines. Plugging in 2 into the first equation can generate our first point:

y = 2(2) - 3 = 4 -3 = 1 this gives us the point (2, 1 )

We can find the distance between this point and the other line by putting the second line into the form ax+by+c=0 :

y = 2x+1 subtract the whole right side from both sides

- 2x + y - 1 = 0 now we see that a = -2, b = 1, c = -1

We can plug the coefficients and the point into the formula

$distance = \frac{|ax + by+c|}{\sqrt{a^2 + b^2 }}$ where (x,y) represents the point.

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

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Example Question #91 : Graphing Functions

Find the distance between $y = \frac{1}5 x +4$ and $y = \frac{1}{5}x - 3$

Possible Answers:

$\frac{7}{\sqrt{29}}$

$\frac{5}{\sqrt{29}}$

$\frac{35}{\sqrt{26}}$

$\frac{7}{\sqrt{26}}$

$\frac{5}{\sqrt{26}}$

Correct answer:

$\frac{35}{\sqrt{26}}$

Explanation:

To find the distance, choose any point on one of the lines. Plugging in into the second equation can generate our first point:

$y = \frac{1}{5}(5) - 3 = 1 -3 = -2$ this gives us the point (5, -2)

We can find the distance between this point and the other line by putting the second line into the form ax+by+c=0 :

$y = \frac{1}{5}x+4$ subtract the whole right side from both sides

$- \frac{1}{5}x + y - 4 = 0$ multiply both sides by

-x + 5y-20=0

now we see that a = -1, b = 5, c = -20

We can plug the coefficients and the point into the formula

$distance = \frac{|ax + by+c|}{\sqrt{a^2 + b^2 }}$ where (x,y) represents the point.

$\frac{|-1(5) + 5(-2)-20| }{ \sqrt{1^2 + 5^2 }} = \frac{|-5-10-20|}{\sqrt{1+25}} = \frac{35}{\sqrt{26}}$

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Example Question #11 : Distance

How far apart are the lines $y = -\frac{3}{2}x -3$ and $y = -\frac{3}{2}x + 6$ ?

Possible Answers:

$\frac{6}{\sqrt{13}}$

$\frac{26}{\sqrt{13}}$

$\frac{18}{\sqrt{13}}$

$\frac{30}{\sqrt{40}}$

$\frac{30}{\sqrt{13}}$

Correct answer:

$\frac{18}{\sqrt{13}}$

Explanation:

To find the distance, choose any point on one of the lines. Plugging in into the first equation can generate our first point:

$y = -\frac{3}{2}(2) - 3 = -3 - 3 =-6$ this gives us the point (2, -6)

We can find the distance between this point and the other line by putting the second line into the form ax+by+c=0 :

$y =- \frac{3}{2}x+6$ subtract the whole right side from both sides

$\frac{3}{2}x + y - 6 = 0$ multiply both sides by

3x +2y - 12 = 0

now we see that a = 2, b = 3, c = -12

We can plug the coefficients and the point into the formula

$distance = \frac{|ax + by+c|}{\sqrt{a^2 + b^2 }}$ where (x,y) represents the point.

$\frac{|3(2) + 2(-6)-12| }{ \sqrt{3^2 + 2^2 }} = \frac{|6-12-12|}{\sqrt{9+4}} = \frac{18}{\sqrt{13}}$

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Example Question #31 : Calculus

Find the distance between $y = \frac{4}{3}x +12$ and $y = \frac{4}{3}x + 2$

Possible Answers:

$\frac{51}{\sqrt{45}}$

$\frac{30}{\sqrt{45}}$

$\frac{51}{5}$

Correct answer:

Explanation:

To find the distance, choose any point on one of the lines. Plugging in into the second equation can generate our first point:

$y = \frac{4}{3}(3) + 2 = 4 + 2 = 6$ this gives us the point (3,6)

We can find the distance between this point and the other line by putting the second line into the form ax+by+c=0 :

$y = \frac{4}{3}x + 12$ subtract the whole right side from both sides

$- \frac{4}{3}x + y - 12 = 0$ multiply both sides by

-4x + 3 y -36 = 0

now we see that a = -4, b = 3, c = -36

We can plug the coefficients and the point into the formula

$distance = \frac{|ax + by+c|}{\sqrt{a^2 + b^2 }}$ where (x,y) represents the point.

$\frac{|-4(3) +3(6)-36| }{ \sqrt{(-4)^2 + 3^2 }} = \frac{|-12+18-36|}{\sqrt{16+9}} = \frac{30}{\sqrt{25}} = \frac{30}{5} = 6$

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Example Question #2 : Find The Distance Between Two Parallel Lines

Find the distance between $y = -\frac{1}{2}x + 7$ and $y = -\frac{1}{2}x + 4$

Possible Answers:

$\frac{42}{\sqrt{34}}$

$\frac{18}{\sqrt{34}}$

$\frac{2}{\sqrt{5}}$

$\frac{9}{\sqrt{34}}$

$\frac{9}{\sqrt{5}}$

Correct answer:

$\frac{2}{\sqrt{5}}$

Explanation:

To find the distance, choose any point on one of the lines. Plugging in into the first equation can generate our first point:

$y = -\frac{1}{2}(4)+7 = -2 + 7 = 5$ this gives us the point (4,5)

We can find the distance between this point and the other line by putting the second line into the form ax+by+c=0 :

$y =-\frac{1}{2}x + 4$ subtract the whole right side from both sides

$- \frac{1}{2}x + y - 4 = 0$ multiply both sides by

x -2y + 8 = 0

now we see that a = 1, b = -2, c =8

We can plug the coefficients and the point into the formula

$distance = \frac{|ax + by+c|}{\sqrt{a^2 + b^2 }}$ where (x,y) represents the point.

$\frac{|1(4) + -2(5)+8| }{ \sqrt{1^2 + (-2)^2 }} = \frac{|4-10+8|}{\sqrt{5}} = \frac{2}{\sqrt{5}}$

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Example Question #31 : Gre Subject Test: Math

Find the distance between the lines $y = \frac{1}{2} x - 11$ and $y = \frac{1}{2} x + 1$

Possible Answers:

$\frac{24}{\sqrt5}$

$\frac{30}{\sqrt5}$

$\frac{24}{\sqrt{52}}$

Correct answer:

$\frac{24}{\sqrt5}$

Explanation:

To find the distance, choose any point on one of the lines. Plugging in into the first equation can generate our first point:

$y = \frac{1}{2}(6) +1 = 3+1= 4$ this gives us the point (6,4)

We can find the distance between this point and the other line by putting the second line into the form ax+by+c=0 :

$y = \frac{1}{2} x - 11$ subtract the whole right side from both sides

$-\frac{1}{2} x + y + 11 = 0$ multiply both sides by

-x + 2y + 22 = 0 now we see that a = -1, b = 2, c = 22

We can plug the coefficients and the point into the formula

$distance = \frac{|ax + by+c|}{\sqrt{a^2 + b^2 }}$ where (x,y) represents the point.

$\frac{|-1(6) + 2(4)+22| }{ \sqrt{(-1)^2 + 2^2 }} = \frac{|-6+8+22|}{\sqrt{1+4}} = \frac{24}{\sqrt{5}}$

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Example Question #35 : Calculus

Find the distance between the points (-4,-11) and (12,7) .

Possible Answers:

None of the Above

$2\sqrt{145}$

$\frac {9}{8}$

$4\sqrt{145}$

Correct answer:

$2\sqrt{145}$

Explanation:

Step 1: Let's define the distance formula. The distance between two sets of coordinates can be found by using the equation:
$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2)}$

In the equation, d is the distance. Also, (x_1,y_1) and (x_2,y_2) are the coordinate points.

(x_1,y_1)=(-4,-11) , (x_2,y_2)=(12,7) .

Step 2: Plug in the values for the missing variables into the equation:

$d=\sqrt{(12-(-4))^2+(7-(-11))^2}$

Step 3: Simplify the inside of the square root. Remember that two minus signs next to each other will change to a plus sign.

$d=\sqrt{(12{\color{Blue} +}4)^2+(7{\color{Blue} +}11)^2}$

Step 4: Add up the numbers in the parentheses:

$d=\sqrt{(16)^2+(18)^2}$

Step 5: Evaluate the exponents:

$d=\sqrt{256+324}$

Step 6: Add the numbers under the square root.

$d=\sqrt {580}$

Step 7: Simplify the number inside the square root as much as possible.

Let's divide by 4:

$\frac {580}{4}=145$ . We cannot break down 145 into another perfect square, so it has to go back into the radical. The square root of 4 is 2, and this will go on the outside.

The final answer is $2\sqrt{145}$

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Example Question #31 : Gre Subject Test: Math

Find the distance between the points (6,-11) and (15,-8) .

Possible Answers:

$10\sqrt{3}$

$-3\sqrt{10}$

$3\sqrt{10}$

Correct answer:

$3\sqrt{10}$

Explanation:

Step 1: The distance formula is defined as:

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

Step 2. Identify what (x_1,y_1) and (x_2,y_2) are.

$(x_1,y_1)=(6,-11)\rightarrow x_1=6, y_1=-11$
$(x_2,y_2)=(15,-8)\rightarrow x_2=15, y_2=-8$

Step 3: Substitute each value for its place in the distance formula.

We will get this:

$d=\sqrt{(15-6)^2+(-8-(-11))^2}$

Step 4: Simplify the inside of step 3.

$d=\sqrt{9^2+(-8+11)^2}$

Step 5: Simplify the parentheses:

$d=\sqrt {9^2+3^2}$

Step 6: Evaluate each exponent:

$d=\sqrt{81+9}->d=\sqrt{90}$

Step 7: Reduce to lowest terms:

Divide $\sqrt 90$ by $\sqrt{9}$ :

$\frac{\sqrt{90}}{\sqrt {9}}=\sqrt {10}$

Step 8: Rewrite $\sqrt{90}$

$\sqrt{90}=\sqrt{9}\cdot \sqrt{10}$

$\sqrt {9}=3$

Replace with $\sqrt {9}$ :

$\sqrt{90}=3\sqrt{10}$

The simplified answer to the question is $3\sqrt{10}$

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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 #11 : Coordinate Geometry

Example Question #1 : Find The Distance Between Two Parallel Lines

Example Question #1 : Find The Distance Between Two Parallel Lines

Example Question #91 : Graphing Functions

Example Question #11 : Distance

Example Question #31 : Calculus

Example Question #2 : Find The Distance Between Two Parallel Lines

Example Question #31 : Gre Subject Test: Math

Example Question #35 : Calculus

Example Question #31 : Gre Subject Test: Math

All GRE Subject Test: Math Resources

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