Distance & Midpoint Formulas - GRE Quantitative Reasoning

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Question

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

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

Answer

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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Question

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

Answer

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

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Question

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

Answer

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.

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

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Question

Find the distance between point $(\frac{1}{2},\frac{1}{2})$ to the line $x=-\frac{1}{2}$ .

Answer

Distance cannot be a negative number. The function $x=-\frac{1}{2}$ is a vertical line. Subtract the value of the line to the x-value of the given point to find the distance.

$D=\left | (-\frac{1}{2})-\frac{1}{2} \right | = \left | -1\right |=1$

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Question

Find the distance between point to line .

Answer

The line is vertical covering the first and fourth quadrant on the coordinate plane.

The x-value of is negative one.

Find the perpendicular distance from the point to the line by subtracting the values of the line and the x-value of the point.

Distance cannot be negative.

$\left | 1-(-1)\right |=\left |2 \right |=2$

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Question

Find the distance between the two lines.

$\ y_1=2.65x-13 \ y_2=2.65x+4$

Answer

Since the slope of the two lines are equivalent, we know that the lines are parallel. Therefore, they are separated by a constant distance. We can then find the distance between the two lines by using the formula for the distance from a point to a nonvertical line:

$d=\frac{|Ax_1 +By_1+C|}{\sqrt{A^{2}+B^{2}}}$

First, we need to take one of the line and convert it to standard form.

where $A=2.65,B=-1, \textup{and }C=-13$

Now we can substitute A, B, and C into our distance equation along with a point, $\left ( x_1,y_1\right )$ , from the other line. We can pick any point we want, as long as it is on line . Just plug in a number for x, and solve for y. I will use the y-intercept, where x = 0, because it is easy to calculate:

Now we have a point, , that is on the line . So let's plug our values for $x_1,y_1,A,B,\textup{and }C}$ :

$d=\frac{|(2.65)\cdot (0) +(-1)\cdot (4)+(-13)|}{\sqrt{(2.65)^{2}+(-1)^{2}}}=\frac{|0-4-13|}{\sqrt{(7.0225+1)}}=\frac{|-17|}{\sqrt{8.0225}}$

$=\frac{17}{2.83}=6.002\approx6$

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Question

How far apart are the line $y = \frac{ 1}{2 } x - 3$ and the point ?

Answer

To find the distance, use the formula $distance = \frac{|ax_{0}+by_{0}+c |}{\sqrt{a^2 + b^2 }}$ where the point is and the line is

First, we'll re-write the equation $y = \frac{1}{2} x - 3$ in this form to identify a, b, and c:

$y = \frac{1}{2} x - 3$ subtract half x and add 3 to both sides

$y - \frac{1}{2}x + 3 = 0$ multiply both sides by 2

Now we see that . Plugging these plus into the formula, we get:

$distance = \frac{|-1(2)+2(7)+6 |}{\sqrt{(-1)^2 +( 2)^2 }} = \frac{|-2 + 14+6|}{\sqrt{1+4} } = \frac{18}{\sqrt5}$

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Question

Find the distance between and

Answer

To find the distance, use the formula $distance = \frac{|ax_{0}+by_{0}+c |}{\sqrt{a^2 + b^2 }}$ where the point is and the line is

First, we'll re-write the equation in this form to identify , , and :

subtract and from both sides

Now we see that . Plugging these plus into the formula, we get:

$distance = \frac{|-4(2)+1(-3)-1 |}{\sqrt{(1)^2 +(-4)^2 }} = \frac{|-8 -3-1|}{\sqrt{1+16} } = \frac{12}{\sqrt{17}}$

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Question

How far apart are the line $y = -\frac{2}{3}x + 8$ and the point ?

Answer

To find the distance, use the formula $distance = \frac{|ax_{0}+by_{0}+c |}{\sqrt{a^2 + b^2 }}$ where the point is and the line is

First, we'll re-write the equation $y = -\frac{2}{3} x +8$ in this form to identify a, b, and c:

$y = -\frac{2}{3} x+8$ add $\frac{2}{3}x$ to and subtract 8 from both sides

$y + \frac{2}{3}x -8 = 0$ multiply both sides by 3

Now we see that . Plugging these plus into the formula, we get:

$distance = \frac{|2(1)+3(3)-24 |}{\sqrt{(3)^2 +( 2)^2 }} = \frac{|2 + 9-24|}{\sqrt{9+4} } = \frac{13}{\sqrt{13}}$

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Question

Find the distance between $y = -\frac{4}{5}x - 2$ and .

Answer

To find the distance, use the formula $distance = \frac{|ax_{0}+by_{0}+c |}{\sqrt{a^2 + b^2 }}$ where the point is and the line is

First, we'll re-write the equation $y = -\frac{4}{5} x -2$ in this form to identify , , and :

$y = -\frac{4}{5} x-2$ add $\frac{4}{5}x$ and to both sides

$y + \frac{4}{5}x +2 = 0$ multiply both sides by

Now we see that . Plugging these plus into the formula, we get:

$distance = \frac{|4(-3)+5(4)+10 |}{\sqrt{(4)^2 +(5)^2 }} = \frac{|-12+20+10|}{\sqrt{16+25} } = \frac{18}{\sqrt{41}}$

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Question

Find the distance between $y = \frac{7}{2} x - 14$ and

Answer

To find the distance, use the formula $distance = \frac{|ax_{0}+by_{0}+c |}{\sqrt{a^2 + b^2 }}$ where the point is and the line is

First, we'll re-write the equation $y = \frac{7}{2} x -14$ in this form to identify , , and :

$y = \frac{7}{2} x-14$ subtract $\frac{7}{2}x$ from and add to both sides

$y -\frac{7}{2}x +14 = 0$ multiply both sides by

Now we see that . Plugging these plus into the formula, we get:

$distance = \frac{|-7(-6)+2(2)+28 |}{\sqrt{(-7)^2 +( 2)^2 }} = \frac{|13 + 4+28|}{\sqrt{49+4} } = \frac{45}{\sqrt{53}}$

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Question

Find the distance between $y = \frac{12}{5} x - 3$ and the point

Answer

To find the distance, use the formula $distance = \frac{|ax_{0}+by_{0}+c |}{\sqrt{a^2 + b^2 }}$ where the point is and the line is

First, we'll re-write the equation $y = \frac{12}{5} x-3$ in this form to identify , , and :

$y = \frac{12}{5} x-3$ subtract $\frac{12}{5}x$ from and add to both sides

$y -\frac{12}{5}x +3 = 0$ multiply both sides by

Now we see that . Plugging these plus into the formula, we get:

$distance = \frac{|-12(3)+5(-4)+15 |}{\sqrt{(5)^2 +( 12)^2 }} = \frac{|-36 + -20+15|}{\sqrt{25+144} } = \frac{41}{\sqrt{169}} = \frac{41}{13}$

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Question

Find the distance between and

Answer

To find the distance, choose any point on one of the lines. Plugging in 2 into the first equation can generate our first point:

this gives us the point

We can find the distance between this point and the other line by putting the second line into the form :

subtract the whole right side from both sides

now we see that

We can plug the coefficients and the point into the formula

$distance = \frac{|ax + by+c|}{\sqrt{a^2 + b^2 }}$ where represents the point.

$\frac{|-2(2) + 1(1)-1| }{ \sqrt{1^2 + 2^2 }} = \frac{|-4+1-1|}{\sqrt{1+4}} = \frac{4}{\sqrt{5}}$

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Question

Find the distance between $y = \frac{1}5 x +4$ and $y = \frac{1}{5}x - 3$

Answer

To find the distance, choose any point on one of the lines. Plugging in into the second equation can generate our first point:

$y = \frac{1}{5}(5) - 3 = 1 -3 = -2$ this gives us the point

We can find the distance between this point and the other line by putting the second line into the form :

$y = \frac{1}{5}x+4$ subtract the whole right side from both sides

$- \frac{1}{5}x + y - 4 = 0$ multiply both sides by

now we see that

We can plug the coefficients and the point into the formula

$distance = \frac{|ax + by+c|}{\sqrt{a^2 + b^2 }}$ where represents the point.

$\frac{|-1(5) + 5(-2)-20| }{ \sqrt{1^2 + 5^2 }} = \frac{|-5-10-20|}{\sqrt{1+25}} = \frac{35}{\sqrt{26}}$

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Question

How far apart are the lines $y = -\frac{3}{2}x -3$ and $y = -\frac{3}{2}x + 6$ ?

Answer

To find the distance, choose any point on one of the lines. Plugging in into the first equation can generate our first point:

$y = -\frac{3}{2}(2) - 3 = -3 - 3 =-6$ this gives us the point

We can find the distance between this point and the other line by putting the second line into the form :

$y =- \frac{3}{2}x+6$ subtract the whole right side from both sides

$\frac{3}{2}x + y - 6 = 0$ multiply both sides by

now we see that

We can plug the coefficients and the point into the formula

$distance = \frac{|ax + by+c|}{\sqrt{a^2 + b^2 }}$ where represents the point.

$\frac{|3(2) + 2(-6)-12| }{ \sqrt{3^2 + 2^2 }} = \frac{|6-12-12|}{\sqrt{9+4}} = \frac{18}{\sqrt{13}}$

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Question

Find the distance between $y = \frac{4}{3}x +12$ and $y = \frac{4}{3}x + 2$

Answer

To find the distance, choose any point on one of the lines. Plugging in into the second equation can generate our first point:

$y = \frac{4}{3}(3) + 2 = 4 + 2 = 6$ this gives us the point

We can find the distance between this point and the other line by putting the second line into the form :

$y = \frac{4}{3}x + 12$ subtract the whole right side from both sides

$- \frac{4}{3}x + y - 12 = 0$ multiply both sides by

now we see that

We can plug the coefficients and the point into the formula

$distance = \frac{|ax + by+c|}{\sqrt{a^2 + b^2 }}$ where represents the point.

$\frac{|-4(3) +3(6)-36| }{ \sqrt{(-4)^2 + 3^2 }} = \frac{|-12+18-36|}{\sqrt{16+9}} = \frac{30}{\sqrt{25}} = \frac{30}{5} = 6$

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Question

Find the distance between $y = -\frac{1}{2}x + 7$ and $y = -\frac{1}{2}x + 4$

Answer

To find the distance, choose any point on one of the lines. Plugging in into the first equation can generate our first point:

$y = -\frac{1}{2}(4)+7 = -2 + 7 = 5$ this gives us the point

We can find the distance between this point and the other line by putting the second line into the form :

$y =-\frac{1}{2}x + 4$ subtract the whole right side from both sides

$- \frac{1}{2}x + y - 4 = 0$ multiply both sides by

now we see that

We can plug the coefficients and the point into the formula

$distance = \frac{|ax + by+c|}{\sqrt{a^2 + b^2 }}$ where represents the point.

$\frac{|1(4) + -2(5)+8| }{ \sqrt{1^2 + (-2)^2 }} = \frac{|4-10+8|}{\sqrt{5}} = \frac{2}{\sqrt{5}}$

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Question

Find the distance between the lines $y = \frac{1}{2} x - 11$ and $y = \frac{1}{2} x + 1$

Answer

To find the distance, choose any point on one of the lines. Plugging in into the first equation can generate our first point:

$y = \frac{1}{2}(6) +1 = 3+1= 4$ this gives us the point

We can find the distance between this point and the other line by putting the second line into the form :

$y = \frac{1}{2} x - 11$ subtract the whole right side from both sides

$-\frac{1}{2} x + y + 11 = 0$ multiply both sides by

now we see that

We can plug the coefficients and the point into the formula

$distance = \frac{|ax + by+c|}{\sqrt{a^2 + b^2 }}$ where represents the point.

$\frac{|-1(6) + 2(4)+22| }{ \sqrt{(-1)^2 + 2^2 }} = \frac{|-6+8+22|}{\sqrt{1+4}} = \frac{24}{\sqrt{5}}$

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Question

Find the distance between the points and .

Answer

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 $2\sqrt{145}$

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Question

Find the distance between the points and .

Answer

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 $3\sqrt{10}$

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