Lines - GMAT Quantitative

Card 0 of 912

Question

Two perpendicular lines intersect at point . One line passes through ; the other, through . What is the value of ?

Answer

The slope of the first line, in terms of , is

$m_{1} =\frac{Y-7}{5-3} = \frac{Y-7}{2}$

The slope of the second line is

$m_{2} =\frac{Y-4}{5-10} = \frac{Y-4}{-5}$

The slopes of two perpendicular lines have product , so we set up this equation and solve for :

$m_{1}m_{2} = 1$

$\frac{Y-7}{2} \cdot \frac{Y-4}{-5} = -1$

$\frac{ \left ( Y-7 \right ) \left ( Y-4 \right ) }{-10} = -1$

$\left ( Y-7 \right ) \left ( Y-4 \right ) = 10$

$Y ^{2}-11Y + 28 = 10$

$Y ^{2}-11Y + 18 = 0$

$\left ( Y - 2 \right ) \left ( Y - 9 \right ) = 0$

$Y - 2 = 0 \Rightarrow Y = 2$

or

$Y - 9 = 0 \Rightarrow Y = 9$

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Question

Which of the following choices give the slopes of two perpendicular lines?

Answer

We can eliminate the choice immediately since the slopes of two perpendicular lines cannot have the same sign. We can also eliminate and undefined, , since a line with slope 0 and a line with undefined slope are perpendicular to each other, not a line of slope -1 or 1.

Of the two remaining choices, we check for the choice that includes two numbers whose product is -1.

$-1.75 \times 1.75 = 3.0625$ and $-0.8 \times 1.25 = -1$

so is the correct choice.

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Question

Which of the following is perpendicular to the line given by the equation:

$y=\frac{2}{3}x-11$

Answer

In order for one line to be perpendicular to another, its slope must be the negative reciprocal of that line's slope. That is, the slope of any perpendicular line must be opposite in sign and the inverse of the slope of the line to which it is perpendicular:

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

In the given line we can see that $m=\frac{2}{3}$ , so the slope of any line perpendicular to it will be:

$m_{perp}=-\frac{1}{m}=-\frac{1}{\frac{2}{3}}=-\frac{3}{2}$

There is only one answer choice with this slope, so we know the following line is perpendicular to the line given in the problem:

$y=-\frac{3}{2}x+7$

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Question

What is the slope of any line that is perpendicular to $y=-\frac{1}{2}x+7$ ?

Answer

For a given line defined by the equation with slope , any line perpendicular to has a slope of $-\frac{1}{m}$ , or the negative reciprocal of . Since the slope of the provided line in this instance is $m=-\frac{1}{2}$ , then the slope of any line perpendicular to is $-\frac{1}{m}=-\frac{1}{-\frac{1}{2}}=2$ .

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Question

What is the slope of any line that is perpendicular to ?

Answer

For a given line defined by the equation with slope , any line perpendicular to has a slope of $-\frac{1}{m}$ , or the negative reciprocal of . Since the slope of the provided line in this instance is , then the slope of any line perpendicular to is $-\frac{1}{m}=-\frac{1}{7}$ .

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Question

What is the slope of any line that is perpendicular to $y=\frac{2}{3}x+5$ ?

Answer

For a given line defined by the equation with slope , any line perpendicular to has a slope of $-\frac{1}{m}$ , or the negative reciprocal of . Since the slope of the provided line in this instance is $m=\frac{2}{3}$ , then the slope of any line perpendicular to is $-\frac{1}{m}=-\frac{1}{\frac{2}{3}}=-\frac{3}{2}$ .

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Question

Find the midpoint of the points and .

Answer

Add the corresponding points together and divide both values by 2:

$(\frac{2+4}{2},\frac{9+3}{2}) = (3, 6)$

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Question

What is the midpoint of and ?

Answer

Add the x-values and divide by 2, and then add the y-values and divide by 2. Be careful of the negatives!

$(\frac{-5+1}{2},\frac{1+4}{2}) = (-2, \frac{5}{2})$

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Question

Which of the following quadrants can contain the midpoint of a line segment with endpoints and for some nonzero value of ?

Answer

The midpoint of the line segment with endpoints and is $\left ( \frac{0+ (-A) }{2}, \frac{A+ 0 }{2}, \right )$ , or $\left (- \frac{A }{2}, \frac{A }{2}, \right )$

If , then the -coordinate is negative and the -coordinate is positive, so the midpoint is in Quadrant II. If , the reverse is true, so the midpoint is in Quadrant IV.

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Question

The midpoint of a line segment with endpoints and is . Sove for .

Answer

The midpoint of a line segment with endpoints $(x_{1}, y_{1}), (x_{2}, y_{2})$ is

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

Substitute the coordinates of the endpoints, then set each equation to the appropriate midpoint coordinate.

-coordinate: $\frac{(A + 2)+ (B + 4) }{2} = 8$

-coordinate: $\frac{(2B-1)+(2A +1) }{2} = 10$

Simplify each, then solve the system of linear equations in two variables:

$\frac{(A + 2)+ (B + 4) }{2} = 8$

$\frac{(A + B + 6) }{2} = 8$

$\frac{(2B-1)+(2A +1) }{2} = 10$

$\frac{2A + 2B }{2} = 10$

The two linear equations turn out to be equivalent, meaning that there are infinitely many solutions to the system. Therefore, insufficient information is given to answer the question.

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Question

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

What are the correct coordinates for the midpoint of $\overline{JK}$ ?

Answer

Midpoint formula is as follows:

$\small m=(\frac{x+x'}{2},\frac{y+y'}{2})$

Plug in and calculate:

$\small \small \small m=(\frac{4+144}{2},\frac{75+5}{2})=(74,40)$

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Question

Segment has endpoints of and . If the midpoint of is given by point , what are the coordinates of point ?

Answer

Midpoints can be found using the following:

$midpoint(x,y)=\left(\frac{x+x'}{2},\frac{y+y'}{2}\right)$

Plug in our points (-6,8) and (4,26) to find the midpoint.

$\small \small midpoint(x,y)=\left(\frac{-6+4}{2},\frac{8+26}{2}\right)=\left(\frac{-2}{2},\frac{34}{2}\right)=(-1,17)$

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Question

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

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

What is the slope of the line perpendicular to ?

Answer

Perpendicular lines have slopes that are negative reciprocals of each other. Therefore, rewrite the equation in slope intercept form :

Slope of given line:

Negative reciprocal:

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Question

What is the slope of a line perpendicular to the line of the equation ?

Answer

The graph of for any real number is a horizontal line. A line parallel to it is a vertical line, which has a slope that is undefined.

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Question

Line 1 is the line of the equation . Line 2 is perpendicular to this line. What is the slope of Line 2?

Answer

Rewrite in slope-intercept form:

The slope of the line is the coefficient of , which is . A line perpendicular to this has as its slope the opposite of the reciprocal of :

$m = -\frac{1}{-1} = 1$

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Question

Given:

$\small f(x)=\frac{5}{6}x-17$

Calculate the slope of , a line perpendicular to .

Answer

To find the slope of a line perpendicular to a given line, simply take the opposite reciprocal of the slope of the given line.

Since f(x) is given in slope intercept form,

$y=mx+b \rightarrow m=slope$ .

Therefore our original slope is

$m=\small \frac{5}{6}$

So our new slope becomes:

$m_{new}=-\frac{1}{m}=-\frac{1}{\frac{5}{6}}=\small -\frac{6}{5}$

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Question

What would be the slope of a line perpendicular to the following line?

Answer

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

Where is the slope of the line and is the y intercept. By definition, the slope of a line is the negative reciprocal of the slope of the line to which it is perpendicular. So if the given line has a slope of , the slope of any line perpendicular to it will have the negative reciprocal of that slope. This gives us:

$y=-5x-3\rightarrow m=-5$

$m_{perp}=-\frac{1}{m}=-\frac{1}{(-5)}=\frac{1}{5}$

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Question

Give the slope of a line on the coordinate plane.

Statement 1: The line shares an -intercept and its -intercept with the line of the equation .

Statement 2: The line is perpendicular to the line of the equation .

Answer

Assume Statement 1 alone. In order to determine the slope of a line on the coordinate plane, the coordinates of two of its points are needed. The -intercept of the line of the equation can be found by substituting and solving for :

The -intercept of the line is at the origin, . It follows that the -intercept is also at the origin. Therefore, Statement 1 only gives one point on the line, and its slope cannot be determined.

Assume Statement 2 alone. The slope of the line of the equation can be calculated by putting it in slope-intercept form :

$7y\div 7 = -5x \div 7$

$y = -\frac{5}{7} x$

The slope of this line is the coefficient of , which is $-\frac{5}{7}$ . A line perpendicular to this one has as its slope the opposite of the reciprocal of $-\frac{5}{7}$ , which is

$- \frac{1}{-\frac{5}{7}} = \frac{7}{5}$ .

The question is answered.

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Question

What is the distance between the points and ?

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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