Coordinate Geometry - GMAT Quantitative

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Question

What are the coordinates of the mipdpoint of the line segment if and

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Answer

The midpoint formula is $(\frac{x_{2}+x_{1}}{2},\frac{y_{2}+y_{1}}{2}) = (\frac{2+2}{2},\frac{3+-7}{2})=(2,-2).$

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Question

A given line has a slope of . What is the slope of any line perpendicular to ?

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Answer

In order for a line to be perpendicular to another line defined by the equation , the slope of line must be a negative reciprocal of the slope of line . Since line 's slope is in the slope-intercept equation above, line 's slope would therefore be $-\frac{1}{m}$ .

Given that we have a line with a slope , we can therefore conclude that any perpendicular line would have a slope $-\frac{1}{m}=-\frac{1}{6}$ .

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Question

Which of the following lines are perpendicular to $y=\frac{5}{6}x+10$ ?

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Answer

In order for a line to be perpendicular to another line defined by the equation , the slope of line must be a negative reciprocal of the slope of line . Since line 's slope is in the slope-intercept equation above, line 's slope would therefore be $-\frac{1}{m}$ .

Since in this instance the slope $m=\frac{5}{6}$ , $-\frac{1}{m}=-\frac{1}{\frac{5}{6}}=-\frac{6}{5}$ . Two of the above answers have this as their slope, so therefore that is the answer to our question.

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Question

Find the slope of a line that is perpendicular to the line running through the points and .

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Answer

To find the slope of the line running through and , we use the following equation:

$m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\frac{8-4}{5-7}=\frac{4}{-2}=-2$

The slope of any line perpendicular to the given line would have a slope that is the negative reciprocal of , or $-\frac{1}{m}$ . Therefore, $-\frac{1}{-2}=\frac{1}{2}$

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Question

Which of the following lines is perpendicular to $y=\frac{3}{4}x+16$ ?

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Answer

Given a line defined by the equation with a slope of , any line perpendicular to would have a slope that is the negative reciprocal of , $-\frac{1}{m}$ . Given our equation $y=\frac{3}{4}x+16$ , we know that $m=\frac{3}{4}$ and that $-\frac{1}{m}=-\frac{1}{\frac{3}{4}}=-\frac{4}{3}$ .

The only answer choice with this slope is $y=-\frac{4}{3}x+12$ .

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Question

Which of the following lines is perpendicular to

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Answer

Given a line defined by the equation with a slope of , any line perpendicular to would have a slope that is the negative reciprocal of , $-\frac{1}{m}$ . Given our equation , we know that and that $-\frac{1}{m}=-\frac{1}{5}$ .

There are two answer choices with this slope, $y=-\frac{1}{5}x-4$ and $y=-\frac{1}{5}x+7$ .

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Question

Do the functions and intersect at a ninety-degree angle, and how can you tell?

$\small f(x)=3x+7$

$\small g(x)=-\frac{1}{3}x+7$

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Answer

If two lines intersect at a ninety-degree angle, they are said to be perpendicular. Two lines are perpendicular if their slopes are opposite reciprocals. In this case:

$f(x):slope=\frac{3}{1}$

$g(x):slope=-\frac{1}{3}$

The two lines' slopes are reciprocals with opposing signs, so the answer is yes. Of our two yes answers, only one has the right explanation. Eliminate the option dealing with -intercepts.

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Question

A given line is defined by the equation . Which of the following lines would be perpendicular to line ?

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Answer

For any line with an equation and slope , a line that is perpendicular to must have a slope of $-\frac{1}{m}$ , or the negative reciprocal of . Given , we know that and therefore know that $-\frac{1}{m}=-\frac{1}{2}$ .

Only one equation above has a slope of $-\frac{1}{2}$ : $y=-\frac{1}{2}x+7$ .

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Question

What is the slope of a line that is perpendicular to $y=-\frac{1}{7}x+9$

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Answer

For any line with an equation and slope , a line that is perpendicular to must have a slope of $-\frac{1}{m}$ , or the negative reciprocal of . Given the equation $y=-\frac{1}{7}x+9$ , we know that $m=-\frac{1}{7}$ and therefore know that $-\frac{1}{m}=-\frac{1}{-\frac{1}{7}}=7$ .

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Question

Which of the following lines is perpendicular to ?

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Answer

For any line with an equation and slope , a line that is perpendicular to must have a slope of $-\frac{1}{m}$ , or the negative reciprocal of . Given the equation , we know that and therefore know that $-\frac{1}{m}=-\frac{1}{4}$ .

Given a slope of $-\frac{1}{4}$ , we know that there are two solutions provided: $y=-\frac{1}{4}x+5$ and $y=-\frac{1}{4}x-4$ .

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Question

What is the slope of a line perpendicular to that of

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Answer

First, we need to rearrange the equation into slope-intercept form. .

$6y-3x=2->y=\frac{3x+2}{6}.$ Therefore, the slope of this line equals $\frac{1}{2}.$ Perpendicular lines have slope that are the opposite reciprocal, or

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Question

What is the distance between the points and ?

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Answer

Let's plug our coordinates into the distance formula.

$\sqrt{(2-7)^{2}+(5-17)^{2}}= \sqrt{(-5)^{2}+(-12)^{2}} = \sqrt{25+144}= \sqrt{169} = 13$

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Question

What is the distance between the points and ?

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Answer

We need to use the distance formula to calculate the distance between these two points.

$\sqrt{(1-5)^{2}+(4-2)^{2}} = \sqrt{(-4)^{2}+(2)^{2}} =\sqrt{20}=\sqrt{4}\sqrt{5}=2\sqrt{5}$

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Question

A line segement on the coordinate plane has endpoints and . Which of the following expressions is equal to the length of the segment?

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Answer

Apply the distance formula, setting

$x_{ 1} = A + 4, x_{2} = A ,y_{ 1} = B, y_{2} = B+3$ :

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

$d = \sqrt{[A-(A+4)]^{2}+[(B+3)-B)]^{2}}$

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

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Question

What is distance between and ?

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Answer

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

$= \sqrt{(6-1)^2+(7-5)^2}$

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

$= \sqrt{(25+4)}$

$\sqrt{29}$

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Question

Consider segment $\overline{JK}$ which passes through the points $\left ( 4,5\right )$ and $\left ( 144,75\right )$ .

Find the length of segment $\overline{JK}$ .

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Answer

This question requires careful application of distance formula, which is really a modified form of Pythagorean theorem.

$\small d=\sqrt{(x-x')^2+(y-y')^2}$

Plug in everthing and solve:

$\small \small d=\sqrt{(144-4)^2+(75-7)^2}=\sqrt{24500}\approx156.5$

So our answer is 156.6

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Question

What is the length of a line segment that starts at the point and ends at the point ?

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Answer

Using the distance formula for the length of a line between two points, we can plug in the given values and determine the length of the line segment by calculating the distance between the two points:

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

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

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

$d=\sqrt{25}$

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Question

What is the equation of the line that is perpendicular to and goes through point ?

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Answer

Perpendicular lines have slopes that are negative reciprocals of each other.

The slope for the given line is , from , where is the slope. Therefore, the negative reciprocal is $-\frac{1}{2}$ .

$m=-\frac{1}{2}$ and :

$y-y_{1}=m(x-x_{1})$

$y-1=-\frac{1}{2}(x-5)$

$y-1=-\frac{1}{2}x+\frac{5}{2}$

$y=-\frac{1}{2}x+\frac{5}{2}+1$

$y=-\frac{1}{2}x+\frac{5}{2}+\frac{2}{2}$

$y=-\frac{1}{2}x+\frac{7}{2}$

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Question

Write the equation of a line that is perpendicular to $y=\frac{-1}{2}x+4$ and goes through point ?

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Answer

A perpendicular line has a negative reciprocal slope to the given line.

The given line, $y=\frac{-1}{2}x+4$ , has a slope of $-\frac{1}{2}$ , as is the slope in the standard form equation .

Slope of perpendicular line:

Point:

Using the point slope formula, we can solve for the equation:

$y-y_{1}=m(x-x_{1})$

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Question

Determine the equation of a line perpendicular to at the point .

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Answer

The equation of a line in standard form is written as follows:

Where is the slope of the line and is the y intercept. First, we can determine the slope of the perpendicular line using the knowledge that its slope must be the negative reciprocal of the slope of the line to which it is perpendicular. For the given line, we can see that , so the slope of a line perpendicular to it will be the negative reciprocal of that value, which gives us:

$m_{perp}=-\frac{1}{m}=-\frac{1}{3}$

Now that we know the slope of the perpendicular line, we can plug its value into the formula for a line along with the coordinates of the given point, allowing us to calculate the -intercept, :

$4=-\frac{1}{3}(2)+b$

$b=\frac{14}{3}$

We now have the slope and the -intercept of the perpendicular line, which is all we need to write its equation in standard form:

$y=-\frac{1}{3}x+\frac{14}{3}$

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