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Warmup 1. Find the equation of a circle with center (-4,1) and radius 3 2. Find an equation of a circle with center at the origin passing through P(4,

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Presentation on theme: "Warmup 1. Find the equation of a circle with center (-4,1) and radius 3 2. Find an equation of a circle with center at the origin passing through P(4,"— Presentation transcript:

1 Warmup 1. Find the equation of a circle with center (-4,1) and radius 3 2. Find an equation of a circle with center at the origin passing through P(4, -7) Copyright © by Houghton Mifflin Company, Inc. All rights reserved.

2 Answers 1. (x+4)2 + (y-1)2 =9 2. x2 + y2 = 65
Copyright © by Houghton Mifflin Company, Inc. All rights reserved.

3 2.3 Linear Equations in Two Variables
Digital Lesson 2.3 Linear Equations in Two Variables

4 The slope of a line is a number, m, which measures its steepness.
y x 2 -2 m is undefined m = 2 m = 1 2 m = 0 m = - 1 4 Copyright © by Houghton Mifflin Company, Inc. All rights reserved. Slope of a Line

5 The slope of the line passing through the two points (x1, y1) and (x2, y2) is given by the formula
y2 – y1 x2 – x1 m = , (x1 ≠ x2 ). x y (x1, y1) (x2, y2) x2 – x1 y2 – y1 change in y change in x Copyright © by Houghton Mifflin Company, Inc. All rights reserved. Slope Formula

6 Example: Find the slope of the line passing through the points (2, 3) and (4, 5).
y2 – y1 x2 – x1 m = 5 – 3 4 – 2 = = 2 = 1 x y (4, 5) (2, 3) 2 2 Copyright © by Houghton Mifflin Company, Inc. All rights reserved. Example: Find Slope

7 Example: Graph the line y = 2x – 4.
(1, -2) 2 (0, - 4) 1 Copyright © by Houghton Mifflin Company, Inc. All rights reserved. Example: y=mx+b

8 A linear equation written in the form y – y1 = m(x – x1) is in point-slope form.
The graph of this equation is a line with slope m passing through the point (x1, y1). Example: The graph of the equation y – 3 = - (x – 4) is a line of slope m = - passing through the point (4, 3). 1 2 x y 4 8 m = - 1 2 (4, 3) Copyright © by Houghton Mifflin Company, Inc. All rights reserved. Point-Slope Form

9 Example: Point-Slope Form
Example: Write the slope-intercept form for the equation of the line through the point (-2, 5) with a slope of 3. y – y1 = m(x – x1) Point-slope form y – y1 = 3(x – x1) Let m = 3. y – 5 = 3(x – (-2)) Let (x1, y1) = (-2, 5). y – 5 = 3(x + 2) Simplify. y = 3x + 11 Slope-intercept form Copyright © by Houghton Mifflin Company, Inc. All rights reserved. Example: Point-Slope Form

10 Example: Slope-Intercept Form
Example: Write the slope-intercept form for the equation of the line through the points (4, 3) and (-2, 5). 2 1 5 – 3 -2 – 4 = - 6 3 Calculate the slope. m = y – y1 = m(x – x1) Point-slope form Use m = - and the point (4, 3). y – 3 = - (x – 4) 1 3 Slope-intercept form y = - x + 13 3 1 Copyright © by Houghton Mifflin Company, Inc. All rights reserved. Example: Slope-Intercept Form

11 Example: Parallel Lines
Two lines are parallel if they have the same slope. If the lines have slopes m1 and m2, then the lines are parallel whenever m1 = m2. x y y = 2x + 4 (0, 4) Example: The lines y = 2x – 3 and y = 2x + 4 have slopes m1 = 2 and m2 = 2. y = 2x – 3 (0, -3) The lines are parallel. Copyright © by Houghton Mifflin Company, Inc. All rights reserved. Example: Parallel Lines

12 Example: Perpendicular Lines
Two lines are perpendicular if their slopes are negative reciprocals of each other. If two lines have slopes m1 and m2, then the lines are perpendicular whenever x y 1 m1 m2= - or m1m2 = -1. y = 3x – 1 y = x + 4 1 3 Example: The lines y = 3x – 1 and y = - x + 4 have slopes m1 = 3 and m2 = 1 3 (0, 4) (0, -1) The lines are perpendicular. Copyright © by Houghton Mifflin Company, Inc. All rights reserved. Example: Perpendicular Lines


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