Other Lines - GMAT Quantitative

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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

In the -plane, what is the slope of the line with equation $4x+5y=8$ ?

Answer

Put the equation in slope-intercept form to solve for the slope:

$y=mx+b$ , where m is the slope and b is the intercept

Rearrange terms: $5y=-4x+8$

Divide by 5: $y=-\frac{4}{5}x+\frac{8}{5}$

slope = $-\frac{4}{5}$

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Question

What is the slope of the line ?

Answer

Rewrite this equation in slope-intercept form: , where is the slope.

$\frac{2Ay}{2A} = \frac{- Ax + 3A}{2A}$

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

The slope is the coefficient of , which is $-\frac{1}{2}$ .

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Question

What is the slope of the line that contains and ?

Answer

The slope formula is:

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

$(1,4)\ and\ (7,10)$

$m= \frac{7-1}{10-4}$

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

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Question

What is the slope of the line that contains and ?

Answer

The slope formula is:

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

$(1,2)\ and\ (5,2)$

$m=\frac{2-2}{5-1}$

$m=\frac{0}{4}$

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Question

What is the slope of the line that contains and ?

Answer

The slope formula is:

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

$(-2,-1)\ and\ (-8,7)$

$m=\frac{7-(-1)}{-8-(-2)}$

$m=\frac{7+1}{-8+2}$

$m=\frac{8}{-6}$

$m=-\frac{4}{3}$

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Question

Give the slope of the line of the equation: $\frac{}{}$ $\frac{2}{5}x + \frac{6}{7}y = 9$

Answer

Rewrite in the slope-intercept form :

$\frac{2}{5}x + \frac{6}{7}y = 9$

$\frac{2}{5}x - \frac{2}{5}x + \frac{6}{7}y = - \frac{2}{5}x+ 9$

$\frac{6}{7}y = - \frac{2}{5}x+ 9$

$\frac{6}{7}y \cdot \frac{7}{6}=\left ( - \frac{2}{5}x+ 9 \right )\cdot \frac{7}{6}$

$y=- \frac{2}{5}\cdot \frac{7}{6}x+ 9 \cdot \frac{7}{6}$

$y=-\frac{7}{15}x+ \frac{21}{2}$

The slope is the coefficient of , which is $-\frac{7}{15}$

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Question

Give the slope of the line of the equation:

Answer

Rewrite in the slope-intercept form :

$\left (0.4x - 0.5y \right )\cdot 10 = 20 \cdot 10$

$\left (0.4x \right )\cdot 10-\left ( 0.5y \right )\cdot 10 = 200$

$-5y \div (-5)=\left ( -4x+ 200 \right )\div (-5)$

$y=\left ( -4x \right ) \div (-5)+\left ( 200 \right ) \div (-5)$

$y= \frac{4}{5}x-40$

The slope is the coefficient of , which is $\frac{4}{5}$ .

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Question

Give the slope of the line of the equation

Answer

Rewrite in the slope-intercept form :

$\frac{1}{5} \cdot 5y = \frac{1}{5} \cdot \left ( x - 7 \right )$

$y = \frac{1}{5} x - \frac{7}{5} \right )$

The slope is the coefficient of , or $\frac{1}{5}$ .

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Question

A iine goes through points and . What is its slope?

Answer

Substitute in the slope formula:

$m = \frac{y_2-y_1}{x_2-x_1}$

$m = \frac{(2N+6 )- (N+4)}{3-6} = \frac{2N+6 -N-4}{-3}= \frac{N+2}{-3} = \frac{-N-2}{3}$

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Question

Give the slope of the line with the equation $\frac{4}{3} x = \frac{2}{5} y - 10$ .

Answer

Rewrite in slope-intercept form:

$\frac{4}{3} x = \frac{2}{5} y - 10$

$\frac{4}{3} x + 10 = \frac{2}{5} y - 10 + 10$

$\frac{2}{5} y = \frac{4}{3} x + 10$

$\frac{5} {2} \cdot \frac{2}{5} y= \frac{5} {2} \cdot \left ( \frac{4}{3} x + 10 \right )$

$y= \frac{5} {2} \cdot \frac{4}{3} x + \frac{5} {2} \cdot 10$

$y= \frac{10}{3} x +25$

The slope is the coefficient of , which is $\frac{10}{3}$ .

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Question

Fill in the circle with a number so that the graph of the resulting equation has slope 4:

$y = 6x + \bigcirc$

Answer

Once a number is filled in, the equation will be in slope-intercept form

,

so the coefficient of will be the slope of the line of the equation. Regardless of the number that is written in the circle, this coefficient, and the slope, will be 6, so the slope cannot be 4.

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Question

Fill in the circle with a number so that the graph of the resulting equation has slope $\frac{4}{5}$ :

$\bigcirc x+ 7y = 20$

Answer

Let be that missing coefficient. Then the equation can be rewritten as

Put the equation in slope-intercept form:

$\frac{7y }{7}=\frac{ -Ax + 20}{7}$

$y =- \frac{ A }{7}x+ \frac{ 20}{7}$

The coefficient of is the slope, so solve for in the equation

$- \frac{ A }{7} =\frac{4}{5}$

$- \frac{ A }{7} \cdot (-7)=\frac{4}{5} \cdot (-7)$

$A=-\frac{28}{5}$

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Question

Fill in the circle with a number so that the graph of the resulting equation has slope 4:

$5 y = \bigcirc x + 4$

Answer

Let be that missing coefficient. Then the equation can be rewritten as

Put the equation in slope-intercept form:

$\frac{5 y}{5} =\frac{ Ax + 4}{5}$

$y = \frac{A}{5}x + \frac{4}{5}$

The coefficient of is the slope, so solve for in the equation

$\frac{A}{5} = 4$

$\frac{A}{5}\cdot 5 = 4 \cdot 5$

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