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Equations of Lines

Y. X. Equations of Lines. Geometry 3.5. Y. X. At the end of this lesson you will be able to:. Write equations for non-vertical lines. Write equations for horizontal lines. Write equations for vertical lines. Use various forms of linear equations.

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Equations of Lines

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  1. Y X Equations of Lines Geometry 3.5

  2. Y X At the end of this lessonyou will be able to: • Write equations for non-vertical lines. • Write equations for horizontal lines. • Write equations for vertical lines. • Use various forms of linear equations. • Calculate the slope of a line passing through two points.

  3. Y Horizontal change (DX) X Before we begin. • Let’s review some vocabulary. Slope (m): The measure of the steepness of a line; it is the ratio of vertical change (DY) to horizontal change (DX). Vertical change (DY) Slope (m) = Y-intercept (b): The y-coordinate of the point where the graph of a line crosses the y-axis. X-intercept (a): The x-coordinate of the point where the graph of a line crosses the x-axis.

  4. Standard Form • Ax + By = C • Graphing find x and y intercepts • Ex 3x + 2y = 6

  5. Y Y-axis (x,y) (0,b) X-axis X Equations ofNon-vertical Lines. • Let’s look at a line with a y-intercept of b, a slopem and let(x,y) be any point on the line.

  6. Y Y-axis (x,y) (y – b) = (0,b) DX X-axis X Slope Intercept Form • The equation for the non-vertical line is: y = mx + b ( Slope Intercept Form ) Where m is: DY DY m = (x – 0) DX

  7. Y (x2,y2) Y-axis (x1,y1) X-axis X More Equations ofNon-vertical Lines. • Let’s look at a line passing through Point 1 (x1,y1) and Point 2 (x2,y2).

  8. Y (x2,y2) Y-axis (y2 – y1) = DX (x1,y1) X-axis X Point Slope Form • The equation for the non-vertical line is: y – y1 = m(x – x1) ( Point Slope Form ) Where m is: DY m = DY (x2 – x1) DX

  9. Y Y-axis (x,b) (0,b) X-axis X Equations ofHorizontal Lines. • Let’s look at a line with a y-intercept of b, a slopem = 0,andlet(x,b) be any point on the Horizontal line.

  10. Y Y-axis (b – b) (x,b) = (0,b) DX DX DY = 0 X-axis X Horizontal Line • The equation for the horizontal line is still y = mx + b ( Slope Intercept Form ). Where m is: DY m = = 0 (x – 0)

  11. Y Y-axis (x,b) (0,b) X-axis X Horizontal Line • Because the value of m is 0, y = mx + b becomes y = b (A Constant Function)

  12. Y Y-axis (a,y) X-axis (a,0) X Equations ofVertical Lines. • Let’s look at a line with no y-interceptb, an x-intercepta, an undefined slopem,andlet(a,y) be any point on the vertical line.

  13. Y Y-axis (a,y) (y – 0) = DX X-axis (a,0) X Vertical Line • The equation for the vertical line is x = a ( a is the X-Intercept of the line). Because m is: DY m = = Undefined (a – a)

  14. Y Y-axis (a,y) X-axis (a,0) X Vertical Line • Because the value of m is undefined, caused by the division by zero, there is no slope m. x = a becomes the equation x = a(The equation of a vertical line)

  15. Y Y-axis DY = 2 DX = 3 (0,3) DX = 3 X-axis X Example 1: Slope Intercept Form • Find the equation for the line with m = 2/3 and b = 3 Because b = 3 The line will pass through (0,3) Because m = 2/3 DY = 2 The Equation for the line is: y = 2/3 x + 3 Mr Brown Honors Geometry

  16. Y X Slope Intercept Form Practice • Write the equation for the lines using Slope Intercept form and then graph the equation. 1.) m = 3 & b = 3 2.) m = 1/4 & b = -2 Mr Brown Honors Geometry

  17. Y Y-axis (10 – -2) 12 = = = 4 DX 3 (6 – 3) X-axis X Example 2: Point Slope Form • Let’s find the equation for the line passing through the points (3,-2) and (6,10) First, Calculate m : (6,10) DY m = DY (3,-2) DX

  18. Y Y-axis X-axis X Example 2: Point Slope Form • To find the equation for the line passing through the points (3,-2) and (6,10) Next plug it into Point Slope From : y – y1 = m(x – x1) Select one point as P1 : (6,10) Let’s use (3,-2) The Equation becomes: y – -2 = 4(x – 3) DY (3,-2) DX

  19. Y Y-axis y + 2 = 4x – 12 -2 = - 2 X-axis X Example 2: Point Slope Form • Simplify the equation / put it into Slope Intercept Form Distribute on the right side and the equation becomes: y + 2 = 4x – 12 Subtract 2 from both sides gives. (6,10) y = 4x – 14 DY (3,-2) DX

  20. Y X Point Slope Form Practice • Find the equation for the lines passing through the following points using Point Slope form. 1.) (3,2) & ( 8,-2) 2.) (5,3) & ( 7,9)

  21. Y Y-axis (2 – 2) (5,2) = (0,2) DX DX DY = 0 X-axis X Example 3: Horizontal Line • Let’s find the equation for the line passing through the points (0,2) and (5,2) y = mx + b ( Slope Intercept Form ). Where m is: DY m = = 0 (5 – 0)

  22. Y Y-axis (5,2) (0,2) X-axis X Example 3: Horizontal Line • Because the value of m is 0, y = 0x + 2 becomes y = 2(A Constant Function)

  23. Y X Horizontal Line Practice • Find the equation for the lines passing through the following points. 1.) (3,2) & ( 8,2) 2.) (4,3) & ( -2,3)

  24. Y Y-axis (3,7) X-axis (3,0) X Example 4: Vertical Line • Let’s look at a line with no y-interceptb, an x-intercepta, passing through (3,0) and (3,7).

  25. Y Y-axis (3,7) 7 (7 – 0) = = DX X-axis (3,0) X Example 4: Vertical Line • The equation for the vertical line is: x = 3 ( 3 is the X-Intercept of the line). Because m is: DY m = = Undefined 0 (3 – 3)

  26. Y X Vertical Line Practice • Find the equation for the lines passing through the following points. 1.) (3,5) & ( 3,-2) 2.) (4,3) & ( 4,-4)

  27. Graphing Equations Conclusions • What are the similarities you see in the equations for Parallel lines? • What are the similarities you see in the equations for Perpendicular lines? • Record your observations on your sheet.

  28. Horizontal change (DX) Equation Summary Vertical change (DY) • Slope: Slope (m) = Slope-Intercept Form: y = mx + b Point-Slope Form: y – y1 = m(x – x1)

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