Other Lines - GMAT Quantitative

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Question

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

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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 ?

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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 ?

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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 ?

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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 ?

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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$

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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:

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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

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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?

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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$ .

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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:

$y = 6x + \bigcirc$

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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 $\frac{4}{5}$ :

$\bigcirc x+ 7y = 20$

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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:

$5 y = \bigcirc x + 4$

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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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Question

Fill in the square and the circle with two numbers so that the line of resulting equation has slope $-$\frac{3 }{5}$$ :

$\square x + 4y = \bigcirc$

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Answer

Let and be those missing numbers. Then the equation can be rewritten as

Put the equation in slope-intercept form:

$\frac{4y}{4}$ =\frac{-Ax+ C}{4}$

$y =-$\frac{A}{4}$x + $\frac{ C}{4}$$

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

$-$\frac{A}{4}$ = -$\frac{3}{5}$$

$-$\frac{A}{4}$ \cdot (-4) = -$\frac{3}{5}$ \cdot (-4)$

$A = $\frac{12}{5}$$

The number in the circle is irrelevant, so the correct choice is that $\frac{12}{5}$ goes in the square and $\frac{4}{7}$ goes in the circle.

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Question

Examine these two equations.

$\square y = x + 7$

Write a number in the box so that the lines of the two equations will have the same slope.

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Answer

Write the first equation in slope-intercept form:

$\frac{3y}{3}$ = $\frac{5x-9}{3}$

$y = $\frac{5}{3}$x - 3$

The coefficient of , which here is $\frac{5}{3}$ , is the slope of the line.

Now, let be the nuimber in the box, and rewrite the second equation as

Write in slope-intercept form:

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

$y = $\frac{1}{A}$x+\frac{7}{A}$$

The slope is $\frac{1}{A}$ , which is set to $\frac{5}{3}$ :

$\frac{1}{A}$= $\frac{5}{3}$

$A= $\frac{3}{5}$$

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Question

Fill in the circle with a number so that the graph of the resulting equation is a horizontal line:

$y = x + \bigcirc$

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Answer

The equation of a horizontal line takes the form for some value of . Regardless of what is written, the equation cannot take this form.

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Question

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

What is the slope of $\overline{JK}$ ?

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Answer

Slope is found via:

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

Plug in and calculate:

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

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Question

Solve for in the coordinate on line ?

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Answer

To solve for for , we have to plug into the variable of the equation and solve for :

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Question

Solve for in the coordinate on line ?

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Answer

To solve for for , we have to plug 1 into the variable of the equation and solve for :

$x=\frac{1}{2}$$

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Question

Solve for in the coordinate on line ?

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Answer

To solve for for , we have to plug into the variable of the equation and solve for :

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