Distance & Midpoint Formulas

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GRE Quantitative Reasoning › Distance & Midpoint Formulas

Questions 1 - 10

Find the distance from point $\left ( 1,-\frac{1}{2}\right )$ to the line .

$\frac{11}{2}$

$\frac{9}{2}$

$-\frac{11}{2}$

$-\frac{9}{2}$

Explanation

Draw a line that connects the point and intersects the line at a perpendicular angle.

The vertical distance from the point $\left ( 1,-\frac{1}{2}\right )$ to the line will be the difference of the 2 y-values.

The distance can never be negative.

$5-\left(-\frac{1}{2}\right) = \frac{10}{2}+\frac{1}{2}=\frac{11}{2}$

Find the length of the line that connects the points: and .

$4\sqrt{5}$

$-4\sqrt{10}$

$4\sqrt{157}$

$\sqrt{586}$

Explanation

Step 1: Recall the distance formula:

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

Step 2: Find ...

Step 3: Substitute the values into the equation:

$d=\sqrt{(8-4)^2+(15-7)^2}$
Reduce the parentheses:

$d=\sqrt{4^2+8^2}$
Evaluate the exponents and add:

$d=\sqrt{16+64}=\sqrt{80}$

Step 4: Reduce $\sqrt{80}$ into lowest terms...

$\sqrt{80}=\sqrt{4}\cdot \sqrt{4}\cdot \sqrt 5$

Using rule of square roots, multiplying two roots with the same value on the inside just gives me the inside value..

$\sqrt{80}=[\sqrt{4}\cdot \sqrt{4}]\cdot \sqrt{5}$
$\sqrt{80}=4\sqrt{5}$

The length of the line that connects both points is $4\sqrt{5}$ .

Find the distance between the points and .

$2\sqrt{145}$

$4\sqrt{145}$

$\frac {9}{8}$

None of the Above

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, and are the coordinate points.

, .

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 $2\sqrt{145}$

Find the midpoint between and

Explanation

To find the midpoint you must use the equation

$\left (\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2} \right )$ Insert numbers

$\left (\frac{5+9}{2}, \frac{-2+-4}{2} \right )$

Find the distance between the two points

$\sqrt{82}$

$\sqrt{58}$

$\sqrt{93}$

Explanation

You must use the distance formula

$\sqrt{({x}_{2}+{x}_{1})^{2}+({y}_{2}+{y}_{1})^{2}}$ Fill in with the points

$\sqrt{({2}+{3})^{2}+({5}+{-4})^{2}}}$ Solve

$\sqrt{({-1})^{2}+({9})^{2}}$

$\sqrt{{1}+{81}}$

$\sqrt{82}$

Given two points, and , find the midpoint.

Explanation

Step 1: Define midpoint. The midpoint is a point located between two given points.. If I draw a line through these points, I get a straight line

Step 2: The midpoint formula is: $\bigg(\frac {x_1+x_2}{2},\frac {y_1+y_2}{2}\bigg)$

Step 3. Plug in the values:

$\bigg(\frac {1+6}{2},\frac {7+11}{2}\bigg)$

Step 4: Simplify each fraction in Step 3:

$\bigg(\frac {7}{2},\frac {18}{2}\bigg)$

Step 5: Convert each fraction to a decimal from step 4:

The midpoint is

What is the shortest distance between the line and the origin?

$\frac{2\sqrt{10}}{5}$

$\frac{4}{3}$

$\frac{3\sqrt{6}}{5}$

$\frac{7\sqrt{2}}{3}$

Explanation

The shortest distance from a point to a line is always going to be along a path perpendicular to that line. To be perpendicular to our line, we need a slope of $-\frac{1}{3}$ .

To find the equation of our line, we can simply use point-slope form, using the origin, giving us

$y-0=\left(-\frac{1}{3}\right)(x-0)$ which simplifies to $y=-\frac{x}{3}$ .

Now we want to know where this line intersects with our given line. We simply set them equal to each other, giving us $-\frac{x}{3}=3x-4$ .

If we multiply each side by , we get .

We can then add to each side, giving us .

Finally we divide by , giving us $x=\frac{6}{5}$ .

This is the x-coordinate of their intersection. To find the y-coordinate, we plug $\frac{6}{5}$ into $y=-\frac{x}{3}$ , giving us $y=-\frac{2}{5}$ .

Therefore, our point of intersection must be $\left(\frac{6}{5},-\frac{2}{5}\right)$ .

We then use the distance formula $d=\sqrt{(x_{1}-x_{2})^2+(y_{1}-y_{2})^2}$ using $\left(\frac{6}{5},-\frac{2}{5}\right)$ and the origin.

This give us $d=\sqrt{\left(\frac{36}{25}\right)+\left(\frac{4}{25}\right)}=\sqrt{\frac{40}{25}}=\frac{2\sqrt{10}}{5}$ .

What is the distance between and

$5\sqrt{2}$

$5\sqrt{3}$

$2\sqrt{5}$

Explanation

Step 1: Plug in values into the distance formula:

$\sqrt{({11-4})^2+(7-6)^2}$

Step 2: Evaluate the inside...

$\sqrt{7^2+1^2}=\sqrt{50}$

Step 3: Simplify...

$\sqrt{50}=\sqrt{25}\sqrt{2}=5\sqrt2$

Find the distance between the points and .

$3\sqrt{10}$

$-3\sqrt{10}$

$10\sqrt{3}$

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 and 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 $3\sqrt{10}$

Find the minimum distance between the point and the following line:

$f(x)=-\frac{1}{3}x+2$

$2\sqrt{10}$

$4\sqrt{5}$

$\frac{9}{2}$

Explanation

The minimum distance from the point to the line would be found by drawing a segment perpendicular to the line directly to the point. Our first step is to find the equation of the new line that connects the point to the line given in the problem. Because we know this new line is perpendicular to the line we're finding the distance to, we know its slope will be the negative inverse of the line its perpendicular to. So if the line we're finding the distance to is:

$f(x)=-\frac{1}{3}x+2$

Then its slope is -1/3, so the slope of a line perpendicular to it would be 3. Now that we know the slope of the line that will give the shortest distance from the point to the given line, we can plug the coordinates of our point into the formula for a line to get the full equation of the new line:

$-4=3(-2)+b\rightarrow b=2$

Now that we know the equation of our perpendicular line, our next step is to find the point where it intersects the line given in the problem:

$3x+2=-\frac{1}{3}x+2$

$\frac{10}{3}x=0\rightarrow x=0$

So if the lines intersect at x=0, we plug that value into either equation to find the y coordinate of the point where the lines intersect, which is the point on the line closest to the point given in the problem and therefore tells us the location of the minimum distance from the point to the line:

So we now know we want to find the distance between the following two points:

and

Using the following formula for the distance between two points, which we can see is just an application of the Pythagorean Theorem, we can plug in the values of our two points and calculate the shortest distance between the point and line given in the problem:

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

$d=\sqrt{(0-(-2))^2+(2-(-4))^2}=\sqrt{(2)^2+(6)^2}=\sqrt{40}$

Which we can then simplify by factoring the radical:

$\sqrt{40}=\sqrt{4\cdot 10}=2\sqrt{10}$

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