Find the Perpendicular Line y=x-3 that contains the point (2,1)

$y = x - 3$ that contains the point $(2, 1)$

Use the slope-intercept form to find the slope.

The slope-intercept form is $y = m x + b$ , where $m$ is the slope and $b$ is the y-intercept.

$y = m x + b$

Using the slope-intercept form, the slope is $1$ .

$m = 1$

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

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

Simplify $- \frac{1}{1}$ to find the slope of the perpendicular line.

Cancel the common factor of $1$ .

Cancel the common factor.

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

Rewrite the expression.

$m_{perpendicular} = - 1 \cdot 1$

Multiply $- 1$ by $1$ .

$m_{perpendicular} = - 1$

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

Use the slope $- 1$ and a given point $(2, 1)$ to substitute for $x_{1}$ and $y_{1}$ in the point-slope form $y - y_{1} = m (x - x_{1})$ , which is derived from the slope equation $m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}$ .

$y - (1) = - 1 \cdot (x - (2))$

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

$y - 1 = - 1 \cdot (x - 2)$

Solve for $y$ .

Simplify $- 1 \cdot (x - 2)$ .

Rewrite.

$y - 1 = 0 + 0 - 1 \cdot (x - 2)$

Simplify by adding zeros.

$y - 1 = - 1 \cdot (x - 2)$

Apply the distributive property.

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

Simplify the expression.

Rewrite $- 1 x$ as $- x$ .

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

Multiply $- 1$ by $- 2$ .

$y - 1 = - x + 2$

Move all terms not containing $y$ to the right side of the equation.

Add $1$ to both sides of the equation.

$y = - x + 2 + 1$

Add $2$ and $1$ .

$y = - x + 3$

Find the Perpendicular Line y=x-3 that contains the point (2,1)

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