Calculating the equation of a line - GMAT Quantitative Reasoning

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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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