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ALGEBRA II HONORS/GIFTED SECTION 2-4 : MORE ABOUT LINEAR EQUATIONS

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Presentation on theme: "ALGEBRA II HONORS/GIFTED SECTION 2-4 : MORE ABOUT LINEAR EQUATIONS"— Presentation transcript:

1 ALGEBRA II HONORS/GIFTED SECTION 2-4 : MORE ABOUT LINEAR EQUATIONS
@ SECTION 2-4 : MORE ABOUT LINEAR EQUATIONS

2 Show that m is the slope in y = mx + b.
Given (x1, y1) and (x2, y2) are two points on a line. y2 = mx2 + b y1 = mx1 + b y2 – y1 = mx2 – mx1 (subtract) y2 – y1 = m(x2 – x1) (factor) (divide by x2 - x1)

3 FORMULAS TO KNOW : Slope-Intercept Form y = mx + b
Standard Form Ax + By = C no fractions, A > 0 Point-Slope Form y - y1 = m(x - x1) Slope

4 Write the equation of the line in slope-intercept form of the line described.
(x1, y1) (x2, y2) Now, use point-slope form to complete the problem. 1) (2, 7) and (4, 10) First, find the slope. y – y1 = m(x – x1) Note : You can use either point for x1 and y1.

5 2) (-2, -4) and (-5, -10) (Write in Standard Form too)

6 6) Contains (-2, -1) and has a slope of
Graphing calculator exercise for parallel and perpendicular lines.

7 PARALLEL LINES : never intersect. They have the same slope.
PERPENDICULAR LINES : meet at right angles. Their slopes are opposite reciprocals. The product of their slopes is -1 (except for horizontal and vertical lines).

8 Write the equation of the line in slope-intercept form of the line described.
7) Contains (5, 8) and is parallel to y = 7x – 6 Parallel lines have the same slope. So, the slope of the line we want is 7. Now, use point-slope form. y – y1 = m(x – x1) y – 8 = 7x - 35 y – 8 = 7(x – 5) y = 7x - 27

9 Transform to slope-intercept form.
Therefore, from Standard Form : the slope is the y-intercept is

10 ALGEBRA II HONORS/GIFTED - SECTION 2-4 (More About Linear Equations)
8) Contains (-3, 0) and is parallel to 2x + 5y = 9 Using our short cut : m = = y – y1 = m(x – x1)

11 9) Contains (-4, 6) and is perpendicular to
Contains (0, 8) and is perpendicular to x – 2y = 1 11) Is horizontal and contains (-8, 9) 12) Is vertical and contains (4, 6) 13) Has x-intercept 5 and slope

12 Another way to handle perpendicular lines.
Write the equation of the line in Standard Form of the line that contains (6, -3) and is perpendicular to x – 5y = 8. 5x + 3y = c Switch the coefficients of x and y, switch the sign in the middle, put c on the right side. 5(6) + 3(-3) = c Substitute 6 for x and -3 for y. 30 + (-9) = c 5x + 3y = 21 = c

13 Write the equation in Standard Form of the line that contains (-5, -1) and perpendicular to 2x + 7y = 3.

14 We’ll end on a sad note!


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