Calculating the equation of a line - GMAT Quantitative

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Find the equation of the line through the points and .

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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

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

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

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

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

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

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

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

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

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}$$

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 .

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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$

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

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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$

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$ .

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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$ .

Find the equation of the line through the points and .

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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

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

Tap to see back →

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

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

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

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

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

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

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

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}$$

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 .

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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$

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

Tap to see back →

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$

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$ .

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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$ .

Find the equation of the line through the points and .

Tap to see back →

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

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

Tap to see back →

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

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

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

Tap to see back →

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

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

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

Tap to see back →

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

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}$$

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 .

Tap to see back →

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$

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

Tap to see back →

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$