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

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Question

One line has four collinear points in order from left to right A, B, C, D. If AB = 10’, CD was twice as long as AB, and AC = 25’, how long is AD?

Answer

AB = 10 ’

BC = AC – AB = 25’ – 10’ = 15’

CD = 2 * AB = 2 * 10’ = 20 ’

AD = AB + BC + CD = 10’ + 15’ + 20’ = 45’

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Question

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

Answer

Let P1 = (1, 4) and P2 = (5, 1)

Substitute these values into the distance formula:

The distance formula is an application of the Pythagorean Theorem: a2 + b2 = c2

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Question

What is the distance of the line drawn between points (–1,–2) and (–9,4)?

Answer

The answer is 10. Use the distance formula between 2 points, or draw a right triangle with legs length 6 and 8 and use the Pythagorean Theorem.

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Question

What is the distance between the points and ?

Answer

Plug the points into the distance formula and simplify:

distance2 = (_x_2 – _x_1)2 + (_y_2 – _y_1)2 = (7 – 3)2 + (2 – 12)2 = 42 + 102 = 116

distance = √116 = √(4 * 29) = 2√29

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Question

What is the distance between two points $\dpi{100} \small (6,14)$ and $\dpi{100} \small (-6,9)$ ?

Answer

To find the distance between two points such as these, plot them on a graph.

Then, find the distance between the $\dpi{100} \small x$ units of the points, which is 12, and the distance between the $\dpi{100} \small y$ points, which is 5. The $\dpi{100} \small x$ represents the horizontal leg of a right triangle and the $\dpi{100} \small y$ represents the vertial leg of a right triangle. In this case, we have a 5,12,13 right triangle, but the Pythagorean Theorem can be used as well.

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Question

Steven draws a line that is 13 units long. If $(-4,1)$ is one endpoint of the line, which of the following might be the other endpoint?

Answer

The distance formula is $\sqrt{((x_{2}-x_{1})^{2} + (y_{2}-y_{1})^{2})}$ .

Plug in $(-4,1)$ with each of the answer choices and solve.

Plug in $(1,13)$ :

$d=\sqrt{(1-(-4))^2+ (13-1)^2} = \sqrt{5^2+12^2}=\sqrt{25+144}=\sqrt{169}=13$

This is therefore the correct answer choice.

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Question

What is the distance between $(1,3)\ and\ (5,6)$ ?

Answer

Let $P_{1}(1,3)$ and $P_{2}(5,6)$ and use the distance formula:

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

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Question

What is the distance between the point and the origin?

Answer

The distance between 2 points is found using the distance or Pythagorean Theorem. Because values are squared in the formula, distance can never be a negative value.

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Question

What is the distance between $(3,4)$ and $(8,16)$ ?

Answer

The formula for the distance between two points is $d = \sqrt{(x_{2} - x_{1})^2+(y_{2} - y_{1})^2}$ .

Plug in the points:

$d=\sqrt{(8-3)^2+(16-4)^2}$

$d = \sqrt{5^2+12^2} = 13$

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Question

Bill gets in his car and drives north for 30 miles at 40 mph. He then turns west and drives 40 mph for 40 minutes. Finally, he goes directly northeast 40 miles in 25 minutes.

Using the total distance traveled as a straight line ("as the crow flies") and the time spent traveling, which of the following is closest to Bill's average speed?

Answer

Each part of the problem gives you 2 out of the 3 pieces of the rate/time/distance relationship, thus allowing you to find the third (if needed) by using the equation:

$Distance = rate\cdot time$

The problem is otherwise an application of geometry and the distance formula. We need to find the distance between 2 points, but we need to go step-by-step to find out where the final point is. The first two steps are relatively easy to follow. He travels 30 miles north. Now he turns west and travels for 40 minutes at 40 mph. This is $\frac{2}{3}$ of an hour, so we have $40\cdot \frac{2}{3}=\frac{80}{3}\ miles$ . Thus if we are looking at standard Cartesian coordinates (starting at the origin), we are now at the point $\left ( \frac{80}{3},30 \right )$ .

We are now on the last step: 40 miles northeast. We need to decipher this into - and - coordinate changes. To do this, we think of a triangle. Because we are moving directly northeast, this is a 45 degree angle to the horizontal. We can thus imagine a 45-45-90 triangle with a hypotenuse of 40. Now using the relationships on triangles we have:

$40^{2} = x^{2} + x^{2} = 2x^{2}\Rightarrow 40 = \sqrt{2x^{2}} = x\sqrt{2} \Rightarrow x = \frac{40}{\sqrt{2}} = 20\sqrt{2}$

So the final step moves us $20\sqrt{2}$ up and to the right. Moving this way from the point $\left ( \frac{80}{3},30 \right )$ leaves us at:

$\left ( \frac{80}{3}-20\sqrt{2}, 30+20\sqrt{2} \right ) \approx \left ( -1.6, 58.3\right )$

Using the distance formula for this point from the origin gives us a distance of ~58 miles.

Now for the time. We traveled 30 miles at 40 mph. This means we traveled for $\frac{30}{40}$ hours or 45 minutes.

We then traveled 40 mph for 40 minutes. Increasing our total time traveled to 85 minutes.

Finally, we traveled 40 miles in 25 minutes, leaving our total time traveled at 110 minutes. Returning to hours, we have $\frac{110}{60}$ hours or 1.83 hours.

Our final average speed traveled is then:

$\frac{58 miles}{1.83 hours} \approx 31.7 mph$

The closest answer to this value is 32 mph.

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Question

Suppose a line is connected from two points. What is the distance of the line if the line is connected from to ?

Answer

Write the distance formula and substitute the values into the formula.

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

Let and .

Plugging in our coordinates into the distance formula we get the following.

$D=\sqrt{((-1)-3)^2+(6-(-9))^2}= \sqrt{(-4)^2+(15)^2}$

$D= \sqrt{16+225}=\sqrt{241}$

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Question

What is the length of the line between the points and ?

Answer

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 are my two points.

Step 2: We need to identify .

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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Question

Find the Distance of the line shown below:

Answer

The distance formula is $^{\sqrt{(x_{1}-x_{2})^{2}+(y_{1}-y_{2})^{2}}}$ . In the graph shown above the coordinates are and . 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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Question

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

Answer

The distance between two points and is given by the Distance Formula:

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

Let 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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Question

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

Answer

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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Question

One long line segment stretches from to . Within that line segment is another, shorter segment that spans from to . What is the distance between the two points on the shorter line segment?

Answer

The distance between two points and is given by the following formula:

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

Let and let . 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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Question

Find the distance from the center of the given circle to the point .

Answer

Remember that the general equation of a circle with center and radius is .

With this in mind, the center of our circle is . To find the distance from this point to , 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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Question

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

Answer

To solve this problem let's choose two vertices that lie diagonally from one another. Let's choose and .

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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Question

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

Answer

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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Question

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

Answer

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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