Graphing a point - GMAT Quantitative

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The point lies on a line with slope that passes through the point .

What is ?

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We first need to find an equation of the line with a slope of that passes through the point (2, 5).

Now, plug in the point and solve for .

A line goes through points . What is ?

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The slope of the line through and can be found using the slope formula:

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

Set :

$m = $\frac{8-4}{4-1}$ = $\frac{4}{3}$$

We can use this slope and the slope formula; set :

$\frac{N-4}{-3-1}$ = $\frac{4}{3}$

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

$$\frac{N-4}{-4}$\cdot \left( -4\right) = $\frac{4}{3}$ \cdot \left( -4\right)$

$N-4= -$\frac{16}{3}$$

$N-4+4= -$\frac{16}{3}$ +4$

$N= -$\frac{4}{3}$ = -1 $\frac{1}{3}$$

The point lies on a line with a slope that passes through . What is the value of ?

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In order to find the value of , we first need to find the equation for the line with slope that passes through the point .

Plugging in and solving for :

The point lies on a line with a slope $m=\frac{1}{2}$$ that passes through . What is the value of ?

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In order to find the value of , we first need to find the equation for the line with a slope $m=\frac{1}{2}$$ that passes through .

$y-6=\frac{1}{2}$(x-3)$

$y-6=\frac{1}{2}$x-$\frac{3}{2}$$

$y=\frac{1}{2}$x+\frac{9}{2}$$

Plugging in and solving for :

$d=\frac{1}{2}$(2)+\frac{9}{2}$$

$d=\frac{11}{2}$$

The point lies on a line with a slope that passes through . What is the value of ?

Tap to see back →

In order to find the value of , we first need to find the equation for the line with a slope that passes through .

Plugging in and solving for :

The point lies on a line with slope that passes through the point .

What is ?

Tap to see back →

We first need to find an equation of the line with a slope of that passes through the point (2, 5).

Now, plug in the point and solve for .

A line goes through points . What is ?

Tap to see back →

The slope of the line through and can be found using the slope formula:

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

Set :

$m = $\frac{8-4}{4-1}$ = $\frac{4}{3}$$

We can use this slope and the slope formula; set :

$\frac{N-4}{-3-1}$ = $\frac{4}{3}$

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

$$\frac{N-4}{-4}$\cdot \left( -4\right) = $\frac{4}{3}$ \cdot \left( -4\right)$

$N-4= -$\frac{16}{3}$$

$N-4+4= -$\frac{16}{3}$ +4$

$N= -$\frac{4}{3}$ = -1 $\frac{1}{3}$$

The point lies on a line with a slope that passes through . What is the value of ?

Tap to see back →

In order to find the value of , we first need to find the equation for the line with slope that passes through the point .

Plugging in and solving for :

The point lies on a line with a slope $m=\frac{1}{2}$$ that passes through . What is the value of ?

Tap to see back →

In order to find the value of , we first need to find the equation for the line with a slope $m=\frac{1}{2}$$ that passes through .

$y-6=\frac{1}{2}$(x-3)$

$y-6=\frac{1}{2}$x-$\frac{3}{2}$$

$y=\frac{1}{2}$x+\frac{9}{2}$$

Plugging in and solving for :

$d=\frac{1}{2}$(2)+\frac{9}{2}$$

$d=\frac{11}{2}$$

The point lies on a line with a slope that passes through . What is the value of ?

Tap to see back →

In order to find the value of , we first need to find the equation for the line with a slope that passes through .

Plugging in and solving for :

The point lies on a line with slope that passes through the point .

What is ?

Tap to see back →

We first need to find an equation of the line with a slope of that passes through the point (2, 5).

Now, plug in the point and solve for .

A line goes through points . What is ?

Tap to see back →

The slope of the line through and can be found using the slope formula:

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

Set :

$m = $\frac{8-4}{4-1}$ = $\frac{4}{3}$$

We can use this slope and the slope formula; set :

$\frac{N-4}{-3-1}$ = $\frac{4}{3}$

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

$$\frac{N-4}{-4}$\cdot \left( -4\right) = $\frac{4}{3}$ \cdot \left( -4\right)$

$N-4= -$\frac{16}{3}$$

$N-4+4= -$\frac{16}{3}$ +4$

$N= -$\frac{4}{3}$ = -1 $\frac{1}{3}$$

The point lies on a line with a slope that passes through . What is the value of ?

Tap to see back →

In order to find the value of , we first need to find the equation for the line with slope that passes through the point .

Plugging in and solving for :

The point lies on a line with a slope $m=\frac{1}{2}$$ that passes through . What is the value of ?

Tap to see back →

In order to find the value of , we first need to find the equation for the line with a slope $m=\frac{1}{2}$$ that passes through .

$y-6=\frac{1}{2}$(x-3)$

$y-6=\frac{1}{2}$x-$\frac{3}{2}$$

$y=\frac{1}{2}$x+\frac{9}{2}$$

Plugging in and solving for :

$d=\frac{1}{2}$(2)+\frac{9}{2}$$

$d=\frac{11}{2}$$

The point lies on a line with a slope that passes through . What is the value of ?

Tap to see back →

In order to find the value of , we first need to find the equation for the line with a slope that passes through .

Plugging in and solving for :

The point lies on a line with slope that passes through the point .

What is ?

Tap to see back →

We first need to find an equation of the line with a slope of that passes through the point (2, 5).

Now, plug in the point and solve for .

A line goes through points . What is ?

Tap to see back →

The slope of the line through and can be found using the slope formula:

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

Set :

$m = $\frac{8-4}{4-1}$ = $\frac{4}{3}$$

We can use this slope and the slope formula; set :

$\frac{N-4}{-3-1}$ = $\frac{4}{3}$

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

$$\frac{N-4}{-4}$\cdot \left( -4\right) = $\frac{4}{3}$ \cdot \left( -4\right)$

$N-4= -$\frac{16}{3}$$

$N-4+4= -$\frac{16}{3}$ +4$

$N= -$\frac{4}{3}$ = -1 $\frac{1}{3}$$

The point lies on a line with a slope that passes through . What is the value of ?

Tap to see back →

In order to find the value of , we first need to find the equation for the line with slope that passes through the point .

Plugging in and solving for :

The point lies on a line with a slope $m=\frac{1}{2}$$ that passes through . What is the value of ?

Tap to see back →

In order to find the value of , we first need to find the equation for the line with a slope $m=\frac{1}{2}$$ that passes through .

$y-6=\frac{1}{2}$(x-3)$

$y-6=\frac{1}{2}$x-$\frac{3}{2}$$

$y=\frac{1}{2}$x+\frac{9}{2}$$

Plugging in and solving for :

$d=\frac{1}{2}$(2)+\frac{9}{2}$$

$d=\frac{11}{2}$$

The point lies on a line with a slope that passes through . What is the value of ?

Tap to see back →

In order to find the value of , we first need to find the equation for the line with a slope that passes through .

Plugging in and solving for :