Algebra Examples

(5, 0), y + 1 = 2 (x - 3)

$(5, 0), y + 1 = 2 (x - 3)$

Step 1

Solve

y + 1 = 2 (x - 3)

$y + 1 = 2 (x - 3)$ .

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Step 1.1

Simplify

2 (x - 3)

$2 (x - 3)$ .

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Step 1.1.1

Rewrite.

y + 1 = 0 + 0 + 2 (x - 3)

$y + 1 = 0 + 0 + 2 (x - 3)$

Step 1.1.2

Simplify by adding zeros.

y + 1 = 2 (x - 3)

$y + 1 = 2 (x - 3)$

Step 1.1.3

Apply the distributive property.

y + 1 = 2 x + 2 \cdot - 3

$y + 1 = 2 x + 2 \cdot - 3$

Step 1.1.4

Multiply

2

$2$ by

- 3

$- 3$ .

y + 1 = 2 x - 6

$y + 1 = 2 x - 6$

y + 1 = 2 x - 6

$y + 1 = 2 x - 6$

Step 1.2

Move all terms not containing

y

$y$ to the right side of the equation.

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Step 1.2.1

Subtract

1

$1$ from both sides of the equation.

y = 2 x - 6 - 1

$y = 2 x - 6 - 1$

Step 1.2.2

Subtract

1

$1$ from

- 6

$- 6$ .

y = 2 x - 7

$y = 2 x - 7$

y = 2 x - 7

$y = 2 x - 7$

y = 2 x - 7

$y = 2 x - 7$

Step 2

Use the slope-intercept form to find the slope.

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Step 2.1

The slope-intercept form is

y = m x + b

$y = m x + b$ , where

m

$m$ is the slope and

b

$b$ is the y-intercept.

y = m x + b

$y = m x + b$

Step 2.2

Using the slope-intercept form, the slope is

2

$2$ .

m = 2

$m = 2$

m = 2

$m = 2$

Step 3

The equation of a perpendicular line must have a slope that is the negative reciprocal of the original slope.

m_{perpendicular} = - \frac{1}{2}

$m_{perpendicular} = - \frac{1}{2}$

Step 4

Find the equation of the perpendicular line using the point-slope formula.

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Step 4.1

Use the slope

- \frac{1}{2}

$- \frac{1}{2}$ and a given point

(5, 0)

$(5, 0)$ to substitute for

x_{1}

$x_{1}$ and

y_{1}

$y_{1}$ in the point-slope form

y - y_{1} = m (x - x_{1})

$y - y_{1} = m (x - x_{1})$ , which is derived from the slope equation

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

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

y - (0) = - \frac{1}{2} \cdot (x - (5))

$y - (0) = - \frac{1}{2} \cdot (x - (5))$

Step 4.2

Simplify the equation and keep it in point-slope form.

y + 0 = - \frac{1}{2} \cdot (x - 5)

$y + 0 = - \frac{1}{2} \cdot (x - 5)$

y + 0 = - \frac{1}{2} \cdot (x - 5)

$y + 0 = - \frac{1}{2} \cdot (x - 5)$

Step 5

Write in

y = m x + b

$y = m x + b$ form.

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Step 5.1

Solve for

y

$y$ .

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Step 5.1.1

Add

y

$y$ and

0

$0$ .

y = - \frac{1}{2} \cdot (x - 5)

$y = - \frac{1}{2} \cdot (x - 5)$

Step 5.1.2

Simplify

$- \frac{1}{2} \cdot (x - 5)$ .

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Step 5.1.2.1

Apply the distributive property.

$y = - \frac{1}{2} x - \frac{1}{2} \cdot - 5$

Step 5.1.2.2

Combine

$x$ and

$\frac{1}{2}$ .

$y = - \frac{x}{2} - \frac{1}{2} \cdot - 5$

Step 5.1.2.3

Multiply

$- \frac{1}{2} \cdot - 5$ .

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Step 5.1.2.3.1

Multiply

$- 5$ by

$- 1$ .

$y = - \frac{x}{2} + 5 (\frac{1}{2})$

Step 5.1.2.3.2

Combine

$5$ and

$\frac{1}{2}$ .

$y = - \frac{x}{2} + \frac{5}{2}$

Step 5.2

Reorder terms.

$y = - (\frac{1}{2} x) + \frac{5}{2}$

Step 5.3

Remove parentheses.

$y = - \frac{1}{2} x + \frac{5}{2}$

Step 6