Other Lines - GMAT Quantitative

Card 0 of 240

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

Tap to see back →

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

What is the slope of the line ?

Tap to see back →

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}$$ .

What is the slope of the line that contains and ?

Tap to see back →

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$

What is the slope of the line that contains and ?

Tap to see back →

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}$$

What is the slope of the line that contains and ?

Tap to see back →

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}$$

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

Tap to see back →

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}$$

Give the slope of the line of the equation:

Tap to see back →

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 $\frac{4}{5}$ .

Give the slope of the line of the equation

Tap to see back →

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 $\frac{1}{5}$ .

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

Tap to see back →

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}$$

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

Tap to see back →

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 $\frac{10}{3}$ .

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

$y = 6x + \bigcirc$

Tap to see back →

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.

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

$\bigcirc x+ 7y = 20$

Tap to see back →

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}$$

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

$5 y = \bigcirc x + 4$

Tap to see back →

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}$\cdot 5 = 4 \cdot 5$

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

$\square x + 4y = \bigcirc$

Tap to see back →

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.

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.

Tap to see back →

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}$$

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

$y = x + \bigcirc$

Tap to see back →

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

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

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

Tap to see back →

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}$$

Solve for in the coordinate on line ?

Tap to see back →

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

Solve for in the coordinate on line ?

Tap to see back →

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

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

Solve for in the coordinate on line ?

Tap to see back →

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