SAT Math - How to find the equation of a line

Whast line goes through the points and ?

Explanation

Let $P_{1}=(1,3)$ and $P_{2}=(7,5)$

The slope is geven by: $m = (y_{2} - y_{1}) \div (x_{2} - x_{1})$ so

$m=\frac{1}{3}$

Then we use the slope-intercept form of an equation; so

$b=\frac{8}{3}$

And we convert

$y=\frac{1}{3}x+\frac{8}{3}$

to standard form.

Let y = 3_x_ – 6.

At what point does the line above intersect the following:

$2x =\frac{2}{3}y+4$

They do not intersect

They intersect at all points

(0,–1)

(–5,6)

(–3,–3)

Explanation

If we rearrange the second equation it is the same as the first equation. They are the same line.

What is the equation of the line with a negative slope that passes through the y-intercept and one x-intercept of the graph y = –x_2 – 2_x + 8 ?

y = –4_x_ + 8

y = –x + 8

y = –2_x_ + 8

y = –2_x_ + 4

y = –4_x_ + 4

Explanation

In order to find the equation of the line, we need to find two points on the line. We are told that the line passes through the y-intercept and one x-intercept of y = –x_2 – 2_x + 8.

First, let's find the y-intercept, which occurs where x = 0. We can substitute x = 0 into our equation for y.

y = –(0)2 – 2(0) + 8 = 8

The y-intercept occurs at (0,8).

To determine the x-intercepts, we can set y = 0 and solve for x.

0 = –x_2 – 2_x + 8

–x_2 – 2_x + 8 = 0

Multiply both sides by –1 to minimize the number of negative coefficients.

x_2 + 2_x – 8 = 0

We can factor this by thinking of two numbers that multiply to give us –8 and add to give us 2. Those numbers are 4 and –2.

x_2 + 2_x – 8= (x + 4)(x – 2) = 0

Set each factor equal to zero.

x + 4 = 0

Subtract 4.

x = –4

Now set x – 2 = 0. Add 2 to both sides.

x = 2

The x-intercepts are (–4,0) and (2,0).

However, we don't know which x-intercept the line passes through. But, we are told that the line has a negative slope. This means it must pass through (2,0).

The line passes through (0,8) and (2,0).

We can use slope-intercept form to write the equation of the line. According to slope-intercept form, y = mx + b, where m is the slope, and b is the y-intercept. We already know that b = 8, since the y-intercept is at (0,8). Now, all we need is the slope, which we can find by using the following formula:

m = (0 – 8)/(2 – 0) = –8/2 = –4

y = mx + b = –4_x_ + 8

The answer is y = –4_x_ + 8.

Find the equation of a line that passes through the point , and is parallel to the line .

Explanation

Since we want a line that is parallel, we will have the same slope as the line . We can use point slope form to create an equation.

, where is the slope and is a point.

What line goes through the points (1, 3) and (3, 6)?

3x + 5y = 2

2x – 3y = 5

4x – 5y = 4

–3x + 2y = 3

–2x + 2y = 3

Explanation

If P1(1, 3) and P2(3, 6), then calculate the slope by m = rise/run = (y2 – y1)/(x2 – x1) = 3/2

Use the slope and one point to calculate the intercept using y = mx + b

Then convert the slope-intercept form into standard form.

Solve the equation for x and y.

x – y = 26/17

2_x_ + 3_y_ = 2

x = 3

y = 2

x = –18/85

y = 112/85

x = 112/85

y = –18/85

x = 85/112

y = –85/18

Explanation

Straightforward problem that presents two unknowns with two equations. The student will need to deal with the fractions correctly to get this one right. Other than the fraction the problem is solved in the exact same manner as the rest in this set. The graph below illustrates the solution.

Sat_math_165_09

Which line contains the following ordered pairs:

$\left ( -2,4 \right )$ and $\left ( 6,2 \right )$

$\small y=-\frac{1}{4}x+\frac{7}{2}$

$\small y=\frac{1}{4}x+\frac{7}{2}$

$\small y=-x+14$

$\small y=x+14$

Explanation

First, solve for slope.

$\small m=\frac{\Delta y}{\Delta x}=\frac{2-4}{6-(-2)}=\frac{-2}{8}=-\frac{1}{4}$

Then, substitute one of the points into the equation y=mx+b.

$\small 2=(-\frac{1}{4})(6)+b$

$\small 2=(-\frac{3}{2})+b$

$\small b=2+\frac{3}{2}=\frac{7}{2}$

This leaves us with the equation $\small y=-\frac{1}{4}+\frac{7}{2}$

What is the slope-intercept form of $\dpi{100} \small 8x-2y-12=0$ ?

$\dpi{100} \small y=4x-6$

$\dpi{100} \small y=4x+6$

$\dpi{100} \small y=2x-3$

$\dpi{100} \small y=-4x+6$

$\dpi{100} \small y=-2x+3$

Explanation

The slope intercept form states that $\dpi{100} \small y=mx+b$ . In order to convert the equation to the slope intercept form, isolate $\dpi{100} \small y$ on the left side:

$\dpi{100} \small 8x-2y=12$

$\dpi{100} \small -2y=-8x+12$

$\dpi{100} \small y=4x-6$

Which of the following equations does NOT represent a line?

Explanation

The answer is .

A line can only be represented in the form or , for appropriate constants , , and . A graph must have an equation that can be put into one of these forms to be a line.

represents a parabola, not a line. Lines will never contain an term.

What is the equation for a line with endpoints (-1, 4) and (2, -5)?

y = -3x + 1

y = 3x - 1

y = -x - 3

y = x + 3

Explanation

First we need to find the slope. Slope (m) = (y2 - y1)/(x2 - x1). Substituting in our values (-5 - 4)/(2 - (-1)) = -9/3 = -3 so slope = -3. The formula for a line is y = mx +b. We know m = -3 so now we can pick one of the two points, substitute in the values for x and y, and find b. 4 = (-3)(-1) + b so b = 1. Our formula is thus y = -3x + 1

How to find the equation of a line

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