Distance Formula - Math

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Question

What is the length of a line segment with end points and ?

Answer

The formula for the length of line is the distance formula, which is very similar to the Pythagorean theorem.

Plug in the given values and solve for the length.

$\sqrt{61}=\sqrt{l^2}$

$\sqrt{61}=l$

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Question

What would be the length of a line with endpoints at and ?

Answer

The formula for the length of line is the distance formula, which is very similar to the Pythagorean theorem.

Plug in the given values and solve for the length.

$\sqrt{73}=\sqrt{l^2}$

$\sqrt{73}=l$

Compare your answer with the correct one above

Question

If a line has a length of , and the endpoints are and , what is the value of ?

Answer

The formula for the length of a line, l, is the distance formula, which is very similar to the Pythagorean Theorem.

Note that the problem has already given us a value for the length of the line. That means . Plug in all of the given values and solve for the missing term.

Subtract $\small 16$ from both sides.

$\sqrt{y^2}=\sqrt{9}$

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Question

What line goes through the points and ?

Answer

Find the slope between the two points:

$m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\frac{15-5}{5-(-1)}=\frac{10}{6}=\frac{5}{3}$

Next, use the slope-intercept form of the equation:

or $15=\frac{5}{3}(5)+b$ where $b=\frac{20}{3}$

So the equation becomes $y=\frac{5}{3}x+ \frac{20}{3}$ or in standard form .

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Question

What is the length of a line with endpoints at and ?

Answer

The formula for the length of a line is very similiar to the pythagorean theorem:

Plug in our given numbers to solve:

$\sqrt{25}=\sqrt{l^2}$

Compare your answer with the correct one above

Question

What is the length of a line with endpoints at and ?

Answer

The formula for the length of a line is very similiar to the pythagorean theorem:

Plug in our given numbers to solve:

$\sqrt{25}=\sqrt{l^2}$

Compare your answer with the correct one above

Question

What is the distance between points and ?

Answer

Use the distance formula:

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

Plug in the given points:

$d=\sqrt{(12-(-2)))^2+(3-4)^2}$

$d=\sqrt{197}$

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Question

Find the distance between points and .

Answer

Use the distance formula:

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

Substitute the given points into the formula:

$d=\sqrt{(14-2)^2+(-2-(-4))^2}$

$d=\sqrt{148}$

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Question

What is the length of a line with endpoints of and ?

Answer

The distance formula is just a reworking of the Pythagorean theorem: $l=\sqrt{\Delta x^2+\Delta y^2}$

Expand that.

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

Plug in our given values.

$l=\sqrt{(7-2)^2+(9--4)^2}$

$l=\sqrt{(5)^2+(13)^2}$

$l=\sqrt{25+169}$

$l=\sqrt{194}$

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Question

What is the distance between and ?

Answer

Let $P_{1}=(1,5)$ and $P_{2} = (6,17)$ .

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

Substitute in the given points:

$d = \sqrt{(6-1)^{2}+(17-5)^{2}}=\sqrt{(5)^{2}+(12)^{2}}=\sqrt{25+144}=\sqrt{169}=13$

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Question

A line segment has endpoints at and . What is the distance of this segment?

Answer

To find the distance, we use the distance formula: $d^2=\Delta x^2+\Delta y^2$ .

Expand that:

Plug in our given values.

$\sqrt{d^2}=\sqrt{81}$

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Question

What is the length of a line segment with end points and ?

Answer

The formula for the length of line is the distance formula, which is very similar to the Pythagorean theorem.

Plug in the given values and solve for the length.

$\sqrt{61}=\sqrt{l^2}$

$\sqrt{61}=l$

Compare your answer with the correct one above

Question

What would be the length of a line with endpoints at and ?

Answer

The formula for the length of line is the distance formula, which is very similar to the Pythagorean theorem.

Plug in the given values and solve for the length.

$\sqrt{73}=\sqrt{l^2}$

$\sqrt{73}=l$

Compare your answer with the correct one above

Question

If a line has a length of , and the endpoints are and , what is the value of ?

Answer

The formula for the length of a line, l, is the distance formula, which is very similar to the Pythagorean Theorem.

Note that the problem has already given us a value for the length of the line. That means . Plug in all of the given values and solve for the missing term.

Subtract $\small 16$ from both sides.

$\sqrt{y^2}=\sqrt{9}$

Compare your answer with the correct one above

Question

What line goes through the points and ?

Answer

Find the slope between the two points:

$m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\frac{15-5}{5-(-1)}=\frac{10}{6}=\frac{5}{3}$

Next, use the slope-intercept form of the equation:

or $15=\frac{5}{3}(5)+b$ where $b=\frac{20}{3}$

So the equation becomes $y=\frac{5}{3}x+ \frac{20}{3}$ or in standard form .

Compare your answer with the correct one above

Question

What is the length of a line with endpoints at and ?

Answer

The formula for the length of a line is very similiar to the pythagorean theorem:

Plug in our given numbers to solve:

$\sqrt{25}=\sqrt{l^2}$

Compare your answer with the correct one above

Question

What is the length of a line with endpoints at and ?

Answer

The formula for the length of a line is very similiar to the pythagorean theorem:

Plug in our given numbers to solve:

$\sqrt{25}=\sqrt{l^2}$

Compare your answer with the correct one above

Question

What is the distance between points and ?

Answer

Use the distance formula:

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

Plug in the given points:

$d=\sqrt{(12-(-2)))^2+(3-4)^2}$

$d=\sqrt{197}$

Compare your answer with the correct one above

Question

Find the distance between points and .

Answer

Use the distance formula:

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

Substitute the given points into the formula:

$d=\sqrt{(14-2)^2+(-2-(-4))^2}$

$d=\sqrt{148}$

Compare your answer with the correct one above

Question

What is the length of a line with endpoints of and ?

Answer

The distance formula is just a reworking of the Pythagorean theorem: $l=\sqrt{\Delta x^2+\Delta y^2}$

Expand that.

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

Plug in our given values.

$l=\sqrt{(7-2)^2+(9--4)^2}$

$l=\sqrt{(5)^2+(13)^2}$

$l=\sqrt{25+169}$

$l=\sqrt{194}$

Compare your answer with the correct one above

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