How to find length of a line - ISEE Lower Level (grades 5-6) Mathematics Achievement

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Question

What is the distance between the points and ?

Answer

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

Let $P_{1}= (-2, 1)$ and $P_{2}=(10, -4)$ .

Plug these two points into the distance formula:

$d = \sqrt{(10-(-2))^{2}+(-4-1)^{2}}=\sqrt{(12)^{2}+(-5)^{2}}=\sqrt{144+25}=\sqrt{169}=13$

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Question

If a line with a length of 5 starts at the point , which of the following is NOT a possible end point?

Answer

Let $P_{1}=(1,1)$ , and let $P_{2}$ be the end point we are looking for.

Recall the distance formula:

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

Squaring the distance formula gives $d^{2} = (x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}$ .

Plug $P_{1}=(1,1)$ this squared distance formula and solve for $P_{2}$ :

$25 = (x_{2}-1)^{2}+(y_{2}-1)^{2}$

Substitute each response into this equation to see which one is false. The only response that doesn't work is .

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Question

What is the distance between and ?

Answer

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

Let $P_{1}= (-2, 1)$ and $P_{2}=(1, 5)$ .

Plug these points into the distance formula:

$d = \sqrt{(1-(-2))^{2}+(5-1)^{2}}=\sqrt{(3)^{2}+(4)^{2}}=\sqrt{9+16}=\sqrt{25}=5$

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Question

If a line that is 13 inches in length starts at the point , which of the following is NOT a possible end point?

Answer

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

Square both sides:

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

Now, let $P_{1}=(1,2)$ , and let $P_{2}$ be the end point we are looking for.

Plug $P_{1}=(1,2)$ into the squared distance formula and solve for $P_{2}$ :

$169 = (x_{2}-1)^{2}+(y_{2}-2)^{2}$

Alternatively, you can plug in the answer choices and see which point makes the above equation untrue. The point matches this description and is therefore the correct answer.

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Question

What is the distance between and ?

Answer

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

Let $P_{1}=(4,-5)$ and $P_{2}=(-2,3)$ :

$d=\sqrt{(-2-4)^{2} + (3-(-5)^{2}}=\sqrt{(-6)^{2} + (8)^{2}}=\sqrt{36 + 64}=\sqrt{100}=10$

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Question

What is the distance between and ?

Answer

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

Let $P_{1}=(1,-3)$ and $P_{2}=(13,-8)$ :

$d=\sqrt{(13-1)^{2} + (-8-(-3)^{2}}=\sqrt{(12)^{2} + (-5)^{2}}=\sqrt{144 + 25}=\sqrt{169}=13$

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Question

What is the distance between and ?

Answer

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

Let $P_{1}=(-7,17)$ and $P_{2}=(2,5)$ :

$d=\sqrt{(2-(-7))^{2} + (5-17)^{2}}=\sqrt{(9)^{2} + (-12)^{2}}=\sqrt{81 + 144}=\sqrt{225}=15$

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Question

How long is a line segment on the number line if it goes from to ?

Answer

To find the length of a line on the number line when one point is negative and the other is positive, you must the the distances of both from zero. is spaces from zero where is the other point is just . So the distance is .

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Question

What is the distance between and

Answer

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

Let $P_{1}=(-7,17)$ and $P_{2}=(2,5)$ :

$d=\sqrt{(2-(-7))^{2} + (5-17)^{2}}=\sqrt{(9)^{2} + (-12)^{2}}=\sqrt{81 + 144}=\sqrt{225}=15$

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