### About Equations of Lines:

We often need to write the equation of a line in different forms. The slope-intercept form can be obtained by solving a linear equation in two variables for y. This gives us y = mx + b, where m is the slope and the y-intercept occurs at (0,b). In some cases, we will not be given enough information to immediately put a line in slope-intercept form. For these scenarios, we are often given a slope and a point on the line or two points on the line and no slope. When this occurs, we can use the point-slope form. This form y - y_{1} = m(x - x_{1}) allows us to plug in the known point for (x_{1},y_{1}) and our known slope m and obtain our slope-intercept form by solving for y. Lastly, we will run into standard form. With standard form, the definition varies from textbook to textbook. Essentially, we see standard form as: ax + by = c, where a, b, and c are integers and a is non-negative. Again this could be relaxed to say a, b, and c are just real numbers. When working with an equation in standard form, we can see that the slope occurs at: m = -a/b and our y-intercept occurs at: y-int: (0, c/b).

Test Objectives

- Demonstrate the ability to write the equation of a line in slope-intercept form
- Demonstrate the ability to write the equation of a line in point-slope form
- Demonstrate the ability to write the equation of a line in standard form

#1:

Instructions: write in slope-intercept form.

$$a)\hspace{.2em}m=\frac{1}{2}, (0,-3)$$

$$b)\hspace{.2em}m=0, (0,-4)$$

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#2:

Instructions: write in slope-intercept form.

$$a)\hspace{.2em}m=-2, (1,0)$$

$$b)\hspace{.2em}m=\frac{3}{2}, (2,4)$$

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#3:

Instructions: write in slope-intercept form.

$$a)\hspace{.2em}m=1, (2, -2)$$

$$b)\hspace{.2em}(5,-4), (5, 3)$$

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#4:

Instructions: write in standard form.

$$b)\hspace{.2em}(0,3), (5,0)$$

$$a)\hspace{.2em}(0,0), (-1,-3)$$

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#5:

Instructions: write in slope-intercept form.

$$a)\hspace{.2em}(-5,-4),(4,0)$$

$$b)\hspace{.2em}(0,-4),(-3,0)$$

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Written Solutions:

#1:

Solutions:

$$a)\hspace{.2em}y=\frac{1}{2}x - 3$$

$$b)\hspace{.2em}y=-4$$

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#2:

Solutions:

$$a)\hspace{.2em}y=-2x + 2$$

$$b)\hspace{.2em}y=\frac{3}{2}x + 1$$

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#3:

Solutions:

$$a)\hspace{.2em}y=x - 4$$

$$b)\hspace{.2em}x=5$$

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#4:

Solutions:

$$a)\hspace{.2em}3x + 5y=15$$

$$b)\hspace{.2em}3x - y=0$$

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#5:

Solutions:

$$a)\hspace{.2em}4x - 9y=16$$

$$b)\hspace{.2em}4x + 3y=-12$$