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SAT Math : How to find the length of a line with distance formula

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Example Question #11 : Distance Formula

Find the Distance of the line shown below:

Screen shot 2015 10 27 at 3.11.25 pm

Possible Answers:

$\sqrt{13}$

$\sqrt{53}$

$2\sqrt{5}$

Correct answer:

$\sqrt{53}$

Explanation:

The distance formula is $^{\sqrt{(x_{1}-x_{2})^{2}+(y_{1}-y_{2})^{2}}}$ . In the graph shown above the coordinates are (-5,8) and (2,10) . When you plug the coordinates into the equation you get:

$\sqrt{(-5-2)^{2}+(8-10)^{2}}$ , which then simplifies to $\sqrt{(-7)^{2}+(2)^{2}}$

$\sqrt{49+4}=\sqrt{53}$ , because is a prime number there is no need to simplify.

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Example Question #12 : Distance Formula

The endpoints of the diameter of a circle are located at (0,0) and (4, 5). What is the area of the circle?

Possible Answers:

$5\pi$

$9\pi$

$\frac{41\pi}{4}$

$41\pi$

$\frac{5\pi}{4}$

Correct answer:

$\frac{41\pi}{4}$

Explanation:

First, we want to find the value of the diameter of the circle with the given endpoints. We can use the distance formula here:

$d=\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}=\sqrt{(4-0)^2 + (5-0)^2}=\sqrt{16 +25}=\sqrt{41}$

If the diameter is $\sqrt{41}$ , then the radius is half of that, or $\frac{\sqrt{41}}{2}$ .

We can then plug that radius value into the formula for the area of a circle.

$A=\pi r^2 = \pi \cdot (\frac{\sqrt{41}}{2})^2 = \frac{41\pi}{4}$ .

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Example Question #13 : Distance Formula

What is the distance between the origin and the point $(\pi, \frac{\pi}{2})$ ?

Possible Answers:

3.14

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

None of the given answers.

$\frac{\pi}{4}\sqrt{5}$

$\frac{\pi}{2}$

Correct answer:

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

Explanation:

The distance between two points (x_1, y_1) and (x_2, y_2) is given by the Distance Formula:

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

Let (x_1, y_1) = (0,0) and $(x_2, y_2) = (\pi, \frac{\pi}{2})$ . Substitute these values into the Distance Formula.

$d=\sqrt{(\pi - 0)^2+(\frac{\pi}{2} - 0)^2} = \sqrt{\pi^2 + (\frac{\pi}{2})^2}=\sqrt{\pi^2 + \frac{\pi^2}{4}}$

To simplify this square root, find a common denominator between the two terms.

$\sqrt{\pi^2 + \frac{\pi^2}{4}}=\sqrt{\frac{4\pi^2}{4} + \frac{\pi^2}{4}}=\sqrt{\frac{5\pi^2}{4}}$

Both 4 and $\pi^2$ are perfect squares, so we can take their square roots to find

$\sqrt{\frac{5\pi^2}{4}}=\frac{\pi}{2}\sqrt{5}$

The distance between our two points is $\frac{\pi}{2}\sqrt{5}$ .

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Example Question #14 : Distance Formula

One long line segment stretches from (0, 2) to (10, 12) . Within that line segment is another, shorter segment that spans from (3, 5) to (9, 11) . What is the distance between the two points on the shorter line segment?

Possible Answers:

$\sqrt{70}$

$5\sqrt{2}$

$6\sqrt{2}$

8.5

Correct answer:

$6\sqrt{2}$

Explanation:

The distance between two points (x_1, y_1) and (x_2, y_2) is given by the following formula:

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

Let (x_1, y_1) = (3, 5) and let (x_2, y_2) = (9, 11) . When we plug these two coordinates into the equation we get:

$d=\sqrt{(9 - 3)^2 + (11 - 5)^2} = \sqrt{6^2 + 6^2} = \sqrt{72} = \sqrt{2 \cdot 36} = \sqrt{2\cdot 6^2} = 6\sqrt{2}$

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Example Question #161 : Lines

Find the distance from the center of the given circle to the point (7, 9) .

(x-2)^2 + (y-3)^2 = 36

Possible Answers:

$\sqrt{60}$

$\sqrt{61}$

Correct answer:

$\sqrt{61}$

Explanation:

Remember that the general equation of a circle with center (h, k) and radius is (x-h)^2 + (y-k)^2 = r^2 .

With this in mind, the center of our circle is (2, 3) . To find the distance from this point to (7, 9) , we can use the distance formula.

$d= \sqrt{(x_2-x_1)^2 + (y_2-y_2)^2} = d= \sqrt{(7-2)^2 + (9-3)^2} = \sqrt{5^2 + 6^2} = \sqrt{25+36} = \sqrt{61}$

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Example Question #16 : Distance Formula

The following points represent the vertices of a box. Find the length of the box's diagonal.

(3, 1), (1, 7), (7, 7), (7, 1)

Possible Answers:

$3\sqrt{20}$

$2\sqrt{15}$

None of the given answers

Correct answer:

$2\sqrt{15}$

Explanation:

To solve this problem let's choose two vertices that lie diagonally from one another. Let's choose (3, 1) and (7, 7) .

We can plug these two points into the Distance Formula, and that will give us the length of the box's diagonal.

$d= \sqrt{(x_2-x_1)^2 + (y_2-y_2)^2} = \sqrt{(7-3)^2 + (7-1)^2} = \sqrt{4^2 + 6^2} = \sqrt{24+36} = \sqrt{60} = \sqrt{2^2 \cdot 15} = 2\sqrt{15}$

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Example Question #17 : Distance Formula

What is the length of the line between the points (6,3) and (-7,4) ?

Possible Answers:

$\sqrt{171}$

-13.04

13.04

13.66

Correct answer:

13.04

Explanation:

Step 1: We need to recall the distance formula, which helps us calculate the length of a line between the two points.

The formula is: $d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$ , where distance and (x_1,y_1);(x_2,y_2) are my two points.

Step 2: We need to identify x_1,x_2,y_1,y_2 .

x_1=6, x_2=-7, y_1=3, y_2=4

Step 3: Substitute the values in step 2 into the formula:

$\sqrt{(-7-6)^2+(4-3)^2}$

Step 4: Start evaluating the parentheses:

$\sqrt {(-13)^2+1^2}$

Step 5: Evaluate the exponents inside the square root

$\sqrt{169+1}$

Step 6: Add the inside:

$\sqrt{170}$

Step 7: We need to evaluate $\sqrt{170}$ in a calculator

$\sqrt{170}\approx 13.038\approx13.04$

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Example Question #11 : Distance Formula

Give the length, in terms of , of a segment on the coordinate plane whose endpoints are (N,N) and $\left ( \frac{1}{2}, \frac{1}{2} \right )$ .

Possible Answers:

$\sqrt{ 2 N^{2} - \frac{1}{2} }$

$\sqrt{ 2 N^{2} + 2 N }$

$\sqrt{ 2 N^{2} - 2 N + \frac{1}{2} }$

$\sqrt{ 2 N^{2} + \frac{1}{2} }$

$\sqrt{ 2 N^{2} + 2 N + \frac{1}{2} }$

Correct answer:

$\sqrt{ 2 N^{2} - 2 N + \frac{1}{2} }$

Explanation:

The length of a segment with endpoints $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ can be calculated using the distance formula:

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

Setting $x_{2} = y_{2} = N$ and $x_{1} = y_{1} = \frac{1}{2}$ and substituting:

$d = \sqrt{ \left (N- \frac{1}{2}\right ) ^{2} + \left (N- \frac{1}{2}\right ) ^{2}}$

The binomials can be rewritten using the perfect square trinomial pattern:

$d = \sqrt{ \left [N^{2} - 2 \cdot N \cdot \frac{1}{2} + \left ( \frac{1}{2}\right ) ^{2} \right ] + \left [N^{2} - 2 \cdot N \cdot \frac{1}{2} + \left ( \frac{1}{2}\right ) ^{2} \right ] }$

$= \sqrt{\left ( N^{2} - N + \frac{1}{4} \right ) +\left ( N^{2} - N + \frac{1}{4} \right ) }$

Simplify and collect like terms:

$d = \sqrt{ N^{2} - N + \frac{1}{4}+ N^{2} - N + \frac{1}{4} }$

$= \sqrt{ N^{2} + N^{2} - N- N + \frac{1}{4} + \frac{1}{4} }$

$= \sqrt{ 2 N^{2} - 2 N + \frac{1}{2} }$

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Example Question #19 : Distance Formula

In terms of , give the length of a segment on the coordinate plane with endpoints $\left (- \frac{1}{4} , N \right )$ and $\left (N, - \frac{1}{4} \right )$ .

Possible Answers:

$\sqrt{2N^{2}+ N}$

$\sqrt{2N^{2}- N}$

$\sqrt{2N^{2}+ \frac{1}{8}}$

$\sqrt{2N^{2}+N+\frac{1}{8}}$

$\sqrt{2N^{2}-N+\frac{1}{8}}$

Correct answer:

$\sqrt{2N^{2}+N+\frac{1}{8}}$

Explanation:

The length of a segment with endpoints $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ can be calculated using the distance formula:

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

Setting $x_{1} = y_{2} = -\frac{1}{4}$ and $x_{2} = y_{1} = N$ , and substituting:

$d = \sqrt{ \left (N - \left (- \frac{1}{4} \right )\right ) ^{2} + \left ( - \frac{1}{4} - N \right ) ^{2}}$

$d = \sqrt{ \left (N + \frac{1}{4} \right ) ^{2} + \left ( - N - \frac{1}{4} \right ) ^{2}}$

The binomials can be rewritten using the perfect square trinomial pattern:

$d = \sqrt{ \left [N^{2}+ 2 \cdot N \cdot \frac{1}{4} + \left ( \frac{1}{4}\right ) ^{2} \right ] + \left [(-N)^{2} - 2 \cdot (-N) \cdot\left ( - \frac{1}{4}\right ) + \left ( - \frac{1}{4}\right ) ^{2} \right ] }$

$d = \sqrt{ \left ( N^{2}+ \frac{1}{2} N + \frac{1}{16} \right ) +\left ( N^{2}+ \frac{1}{2} N + \frac{1}{16} \right ) }$

Simplify and collect like terms:

$d = \sqrt{ N^{2}+ N^{2}+ \frac{1}{2} N + \frac{1}{2} N + \frac{1}{16} + \frac{1}{16} }$

$= \sqrt{2 N^{2}+ N + \frac{1}{8} }$

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Example Question #20 : Distance Formula

Give the length, in terms of , of a line segment on the coordinate plane whose endpoints are (N, N) and $\left ( \frac{1}{3}, -\frac{1}{3} \right )$ .

Possible Answers:

$\sqrt{2N^{2}+\frac{4}{3}N+\frac{2}{9}}$

$\sqrt{2N^{2}+\frac{4}{3}N }$

$\sqrt{2N^{2}+\frac{2}{9}}$

$\sqrt{2N^{2}-\frac{4}{3}N }$

$\sqrt{2N^{2}-\frac{4}{3}N+\frac{2}{9}}$

Correct answer:

$\sqrt{2N^{2}+\frac{2}{9}}$

Explanation:

The length of a segment with endpoints $(x_{1},y_{1}) , (x_{2},y_{2})$ can be calculated using the distance formula

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

Substituting $x_{1} = \frac{1}{3} ,y_{1} =-\frac{1}{3}, x_{2} = y_{2}= N$ , and simplifying"

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

$= \sqrt{\left (N - \frac{1}{3}\right )^{2}+ \left (N - \left ( - \frac{1}{3} \right ) \right )^{2} }$

$= \sqrt{\left (N - \frac{1}{3}\right )^{2}+ \left (N + \frac{1}{3} \right ) ^{2} }$

$= \sqrt{ \left [ N^{2} - 2 \cdot N \cdot \frac{1}{3} + \left ( \frac{1}{3}\right )^{2} \right ]+ \left [ N ^{2} + 2 \cdot N \cdot \frac{1}{3} + \left ( \frac{1}{3}\right )^{2} \right ] }$

$= \sqrt{ \left ( N^{2} - \frac{2}{3}N + \frac{1}{9} \right ) + \left ( N^{2} + \frac{2}{3}N + \frac{1}{9} \right ) }$

$= \sqrt{ N^{2} + N^{2} - \frac{2}{3}N + \frac{2}{3}N + \frac{1}{9} + \frac{1}{9} }$

$= \sqrt{ 2N^{2} +\frac{2} {9} }$

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SAT Math : How to find the length of a line with distance formula

Study concepts, example questions & explanations for SAT Math

All SAT Math Resources

Example Questions

Example Question #11 : Distance Formula

Example Question #12 : Distance Formula

Example Question #13 : Distance Formula

Example Question #14 : Distance Formula

Example Question #161 : Lines

Example Question #16 : Distance Formula

Example Question #17 : Distance Formula

Example Question #11 : Distance Formula

Example Question #19 : Distance Formula

Example Question #20 : Distance Formula

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