Lines - GMAT Quantitative Reasoning

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Question

The midpoint of a line segment with endpoints and is . What is ?

Answer

If the midpoint of a line segment with endpoints and is , then by the midpoint formula,

$\frac{2A + (B + 7)}{2} = 9$

and

$\frac{B + (3A - 2)}{2} = 6$ .

The first equation can be simplified as follows:

or

The second can be simplified as follows:

or

This is a system of linear equations. can be calculated by subtracting:

$\underline{2A + B = 11 }$

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Question

The quadrilateral with vertices is a trapezoid. What are the endpoints of its midsegment?

Answer

The midsegment of a trapezoid is the segment whose endpoints are the midpoints of its legs - its nonparallel opposite sides. These two sides are the ones with endpoints and . The midpoint of each can be found by taking the means of the - and -coordinates:

$(0,0), (4,4): \left ( \frac{0+4}{2},\frac{0+4}{2} \right ) \Rightarrow (2,2)$

$(22,0), (16,4): \left ( \frac{22+16}{2},\frac{0+4}{2} \right ) \Rightarrow (19,2)$

The midsegment is the segment that has endpoints (2,2) and (19,2)

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Question

If the midpoint of $\overline{VU}$ is $\left ( 6,7\right )$ and $\textup{Point }V$ is at $\left (2,9\right )$ , what are the coordinates of $\textup{Point }U$ ?

Answer

Midpoint formula is as follows:

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

In this case, we have x,y and the value of the midpoint. We need to findx' and y'

V is at (2,9) and the midpoint is at (6,7)

$\small 6=\frac{2+x'}{2}$ and $\small 7=\frac{9+y'}{2}$

$\small x'=62-2=10$ $\small y'=72-9=14-9=5$

So we have (10,5) as point U

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Question

A line segment has its midpoint at and an endpoint at . What are the coordinates of the other endpoint?

Answer

Because we are given the midpoint and one of the endpoints, we know the x coordinate of the other endpoint will be the same distance away from the midpoint in the x direction, and the y coordinate of the other endpoint will be the same distance away from the midpoint in the y direction. Given two endpoints of the form:

The midpoint of these two endpoints has the coordinates:

$(x_m,y_m)=(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})$

Plugging in values for the given midpoint and one of the endpoints, which we can see is because it lies to the right of the midpoint, we can solve for the other endpoint as follows:

$(x_m,y_m)=(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})$

$(-1,3)=(\frac{x_1+3}{2},\frac{y_1+8}{2})$

$\frac{x_1+3}{2}=-1\rightarrow x_1+3=-2\rightarrow x_1=-5$

$\frac{y_1+8}{2}=3\rightarrow y_1+8=6\rightarrow y_1=-2$

So the other endpoint has the coordinates

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Question

Consider segment with endpoint at . If the midpoint of can be found at , what are the coordinates of point ?

Answer

Recall midpoint formula:

$(x',y')=\left(\frac{x^1+x^2}{2}, \frac{y^1+y^2}{2}\right)$

In this case we have (x'y') and one of our other (x,y) points.

Plug and chug:

$(-15,-67)=\left(\frac{8+x}{2}, \frac{9+y}{2}\right)$

If you make this into two equations and solve you get the following.

$\small -15=\frac{8+x}{2}, x=-38$

$\small -67=\frac{9+y}{2}, y=-143$

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Question

Find the equation of the line through the points and .

Answer

First find the slope of the equation.

$m =\frac{rise}{run} =\frac{7 + 2}{1-4} = \frac{9}{-3}=-3$

Now plug in one of the two points to form an equation. Here we use (4, -2), but either point will produce the same answer.

$y-(-2)=-3(x-4)$

$y + 2=-3x + 12$

$y=-3x+10$

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Question

What is the equation of a line with slope and a point ?

Answer

Since the slope and a point on the line are given, we can use the point-slope formula:

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

$m=1\ and\ (1,4)$

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Question

What is the equation of a line with slope and point ?

Answer

Since the slope and a point on the line are given, we can use the point-slope formula:

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

$m=0\ and\ (5,2)$

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Question

What is the equation of a line with slope $-\frac{4}{3}$ and a point ?

Answer

Since the slope and a point on the line are given, we can use the point-slope formula:

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

slope: $-\frac{4}{3}$ and point:

$y-7=-\frac{4}{3}(x-(-8))$

$y-7=-\frac{4}{3}(x+8)$

$y-7=-\frac{4}{3}x-\frac{32}{3}$

$y=-\frac{4}{3}x-\frac{32}{3}+7$

$y=-\frac{4}{3}x-\frac{32}{3}+\frac{21}{3}$

$y=-\frac{4}{3}x-\frac{11}{3}$

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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 equation of $\overline{JK}$ in the form .

Answer

Given that JK passes through (4,5) and (144,75) we can find the slope as follows:

Slope is found via:

$m=\frac{y'-y}{x'-x}$

Plug in and calculate:

$\small \small m=\frac{75-5}{144-4}=\frac{70}{140}=\frac{1}{2}$

Next, we need to use one of our points and the slope to find our y-intercept. I'll use (4,5).

$\small 5=\frac{1}{2}*4+b$

$\small b=5-2=3$

So our answer is:

$\small y=\frac{x}{2}+3$

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Question

Determine the equation of a line that has the points and ?

Answer

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

Where is the slope and is the y intercept. We start by calculating the slope between the two given points using the following formula:

$m=\frac{y_2-y_1}{x_2-x_1}=\frac{7-3}{6-(-2)}=\frac{4}{8}=\frac{1}{2}$

Now we can plug either of the given points into the formula for a line with the calculated slope and solve for the y intercept:

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

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

$y=\frac{1}{2}x+4$

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Question

Give the equation of the line that passes through the -intercept and the vertex of the parabola of the equation

$y = x^{2} + 5x - 6$ .

Answer

The -intercept of the parabola of the equation can be found by substituting 0 for :

$y = x^{2} + 5x - 6$

$y = 0^{2} + 5 (0) - 6$

This point is .

The vertex of the parabola of the equation $y = a x^{2} + bx+c$ has -coordinate $- \frac{b}{2a}$ , and its -coordinate can be found using substitution for . Setting and :

$x = - \frac{b}{2a} = - \frac{5}{2 \cdot 1} = - \frac{5}{2 }$

$y = \left ( - \frac{5}{2 } \right )^{2} + 5 \left ( - \frac{5}{2 } \right ) - 6$

$y = \frac{25}{4 } - \frac{25}{2 } - 6$

$y = \frac{25}{4 } - \frac{50}{4 } - \frac{24}{4}$

$y = - \frac{49}{4 }$

The vertex is $\left (- \frac{5}{2 },- \frac{49}{4 } \right )$

The line connects the points and $\left (- \frac{5}{2 },- \frac{49}{4 } \right )$ . Its slope is

$m= \frac{y_{2} - y_{1}}{x_{2} - x_{1}}$

$= \frac{-6-\left ( - \frac{49}{4 } \right )}{0 - \left (- \frac{5}{2 } \right )}$

$= \frac{-\frac{24 }{4 } + \frac{49}{4 } }{ \frac{5}{2 } }$

$= \frac{ \frac{25}{4 } }{ \frac{5}{2 } }$

$= \frac{25}{4 } \cdot \frac{2 } {5}$

$=\frac{5}{2 }$

Since the line has -intercept and slope $m =\frac{5}{2 }$ , the equation of the line is $y=\frac{5}{2 } x + (- 6)$ , or $y=\frac{5}{2 } x - 6$ .

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Question

Find the equation of a line that is parallel to and passes through the point .

Answer

The parallel line has the equation $\dpi{100} \small 4x-2y=5$ . We can find the slope by putting the equation into slope-intercept form, y = mx + b, where m is the slope and b is the intercept. $\dpi{100} \small 4x-2y=5$ becomes $\dpi{100} \small y=2x-\frac{5}{2}$ , so the slope is 2.

We know that our line must have an equation that looks like $\dpi{100} \small y=2x+b$ . Now we need the intercept. We can solve for b by plugging in the point (4, 1).

1 = 2(4) + b

b = –7

Then the line in question is $\dpi{100} \small y=2x-7$ .

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Question

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

Answer

Parallel lines have the same slope. Therefore, the slope of the new line is , as the equation of the original line is ,with slope .

and :

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

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Question

Given:

$\small f(x)=4x+13$

Which of the following is the equation of a line parallel to that has a y-intercept of ?

Answer

Parallel lines have the same slope, so our slope will still be 4. The y-intercept is just the "+b" at the end. In f(x) the y-intercept is 13. In this case, we need to have a y-intercept of -13, so our equation just becomes:

$\small \small f(x)=4x-13$

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Question

Find the equation of the line that is parallel to the and passes through the point .

$\small g(x)=6x+7$

Answer

Two lines are parallel if they have the same slope. The slope of g(x) is 6, so eliminate anything without a slope of 6.

Recall slope intercept form which is .

We know that the line must have an m of 6 and an (x,y) of (8,9). Plug everything in and go from there.

$9=(6\cdot 8)+b$

$\small 9=48+b$

$\small b=9-48=-39$

So we get:

$\small y=6x-39$

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Question

Given the function , which of the following is the equation of a line parallel to and has a -intercept of ?

Answer

Given a line defined by the equation with slope , any line that is parallel to also has a slope of . Since , the slope is and the slope of any line parallel to also has a slope of .

Since also needs to have a -intercept of , then the equation for must be .

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Question

Given the function $f(x)=-\frac{6}{5}x+7$ , which of the following is the equation of a line parallel to and has a -intercept of ?

Answer

Given a line defined by the equation with slope , any line that is parallel to also has a slope of . Since $f(x)=-\frac{6}{5}x+7$ , the slope is $-\frac{6}{5}$ and the slope of any line parallel to also has a slope of $m=-\frac{6}{5}$ .

Since also needs to have a -intercept of , then the equation for must be $g(x)=-\frac{6}{5}x-2$ .

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Question

Given the function , which of the following is the equation of a line parallel to and has a -intercept of ?

Answer

Given a line defined by the equation with slope , any line that is parallel to also has a slope of . Since , the slope is and the slope of any line parallel to also has a slope of .

Since also needs to have a -intercept of , then the equation for must be .

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Question

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

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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