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Working with Parallel and Perpendicular Lines Adapted from Walch Education.

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Presentation on theme: "Working with Parallel and Perpendicular Lines Adapted from Walch Education."— Presentation transcript:

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2 Working with Parallel and Perpendicular Lines Adapted from Walch Education

3 Key Concepts We can write the equation of a line through a given point that is parallel to a given line if you know the equation of the given line. It is necessary to identify the slope of the given equation before trying to write the equation of the line that is parallel or perpendicular. If the given equation is not in slope-intercept form, rewrite it.

4 Writing Equations Parallel to a Given Line Through a Given Point Rewrite the given equation in slope-intercept form if necessary. Identify the slope of the given line. Write the general point-slope form of a linear equation: y – y 1 = m(x – x 1 ). Substitute the slope of the given line for m in the general equation. Substitute x and y from the given point into the general equation for x 1 and y 1. Simplify the equation. Rewrite the equation in slope-intercept form if necessary.

5 Writing Equations Perpendicular to a Given Line Through a Given Point Rewrite the given equation in slope-intercept form if necessary. Identify the slope of the given line. Find the opposite reciprocal of the slope of the given line. Write the general point-slope form of a linear equation: y – y 1 = m(x – x 1 ). Substitute the opposite reciprocal of the given line for m in the general equation. Substitute x and y from the given point into the general equation for x 1 and y 1. Simplify the equation. Rewrite the equation in slope-intercept form if necessary.

6 More Concepts to Know: The shortest distance between two points is a line. The shortest distance between a given point and a given line is the line segment that is perpendicular to the given line through the given point (we will explore this further in class).

7 Practice #1 Write the slope-intercept form of an equation for the line that passes through the point (5, –2) and is parallel to the graph of 8x – 2y = 6.

8 Rewrite the given equation in slope-intercept form. 8x – 2y = 6Given equation –2y = 6 – 8xSubtract 8x from both sides. y = –3 + 4xDivide both sides by –2. y = 4x – 3 Write the equation in slope- intercept form.

9 Substitute the slope of the given line for m in the point-slope form of a linear equation. y – y 1 = m(x – x 1 )Point-slope form y – y 1 = 4(x – x 1 )Substitute m from the given equation. Substitute x and y from the given point into the equation for x 1 and y 1. y – y 1 = 4(x – x 1 ) Equation y – (–2) = 4(x – 5) Substitute (5, –2) for x 1 and y 1.

10 Simplify the equation. y – (–2) = 4(x – 5)Equation with substituted values for x 1 and y 1 y – (–2) = 4x – 20Distribute 4 over (x – 5). y + 2 = 4x – 20Simplify. y = 4x – 22Subtract 2 from both sides. The equation of the line through the point (5, –2) that is parallel to the equation 8x – 2y = 6 is y = 4x – 22.

11 Challenge Find the point on the line y = 4x + 1 that is closest to the point (–2, 8).

12 Thanks for Watching !!!! ~Ms. Dambreville


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