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Algebra 2 More About Linear Equations Lesson 2-4 Part 1 Goals Goal • To write equations of lines. Rubric Level 1 – Know the goals. Level 2 – Fully understand the goals. Level 3 – Use the goals to solve simple problems. Level 4 – Use the goals to solve more advanced problems. Level 5 – Adapts and applies the goals to different and more complex problems. Essential Question Big Idea: Equivalence • Why are there different forms for the equation of a line? Vocabulary • Point-Slope Form • Standard Form Linear Equations Recall from Lesson 2-3 that the slope-intercept form of a linear equation is y= mx + b, where m is the slope of the line and b is its y-intercept. And if given the slope and y-intercept, you can write the equation of a line in slope-intercept form. You can also write the equation of a line given any point on the line and the slope using the point-slope form of a linear equation. Point-Slope Form If you know the slope and any point on the line, you can write an equation of the line by using the slope formula. Point-Slope Form y – y1 = m(x – x1) y-coordinate x-coordinate slope (x1, y1) represents a specific point and (x, y) represents any point. Example: Writing Linear Equations in Point-Slope Form Write an equation in point-slope form for the line with the given slope that contains the given point. A. B. C. y 3 4 x y 4 x 1 Your Turn: Write an equation in point-slope form for the line with the given slope that contains the given point. a. b. slope = 0; (3, –4) y – (–4) = 0(x – 3) y + 4 = 0(x – 3) y40 Example: Writing Linear Equations Using Point-Slope Form Write an equation in slope-intercept form for the line with slope 3 that contains (–1, 4). Step 1 Write the equation in point-slope form: y – y1 = m(x – x1) y – 4 = 3[x – (–1)] Step 2 Rewrite the equation in slope-intercept form by solving for y. y – 4 = 3(x + 1) y – 4 = 3x + 3 +4 +4 y = 3x + 7 Rewrite subtraction of negative numbers as addition. Distribute 3 on the right side. Add 4 to both sides. Your Turn: Write an equation in slope-intercept form for the line with slope that contains (–3, 1). Step 1 Write the equation in point-slope form: y – y1 = m(x – x1) Continued Step 2 Rewrite the equation in slope-intercept form by solving for y. Rewrite subtraction of negative numbers as addition. Distribute +1 +1 on the right side. Add 1 to both sides. Your Turn; Write the equation of the line in slope-intercept form with slope –5 through (1, 3). Point-Slope Form y – y1 = m(x – x1) y – (3) = –5(x – 1) y – 3 = –5(x – 1) Substitute. Simplify. Rewrite in slope-intercept form. y – 3 = –5(x – 1) y – 3 = –5x + 5 y = –5x + 8 Distribute. Solve for y. The equation of the line is y = –5x + 8. Point-Slope Form Point-Slope Form can also be used to write the equation of a line given two points. • Procedure 1) Use the two given points to calculate the slope. 2) Substitute the slope and one of the given points into the point-slope form. 3) Rewrite equation in slope-intercept form if required. Example: Write Equation of Line Given Two Points Write an equation in slope-intercept form for the line through the two points. (2, –3) and (4, 1) Step 1 Find the slope. Step 2 Substitute the slope and one of the points into the pointslope form. y – y1 = m(x – x1) y – (–3) = 2(x – 2) Choose (2, –3). Example: Continued y – (–3) = 2(x – 2) Step 3 Write the equation in slope-intercept form. y + 3 = 2(x – 2) y + 3 = 2x – 4 –3 –3 y = 2x – 7 Example: Write Equation of Line Given Two Points Write an equation in slope-intercept form for the line through the two points. (0, 1) and (–2, 9) Step 1 Find the slope. Step 2 Substitute the slope and one of the points into the pointslope form. y – y1 = m(x – x1) y – 1 = –4(x – 0) Choose (0, 1). Example: Continued y – 1 = –4(x – 0) Step 3 Write the equation in slope-intercept form. y – 1 = –4(x – 0) y – 1 = –4x + 1 +1 y = –4x + 1 Your Turn: Write an equation in slope-intercept form for the line through the two points. (1, –2) and (3, 10) Step 1 Find the slope. Step 2 Substitute the slope and one of the points into the pointslope form. y – y1 = m(x – x1) y – (–2) = 6(x – 1) y + 2 = 6(x – 1) Choose (1, –2). Continued y + 2 = 6(x – 1) Step 3 Write the equation in slope-intercept form. y + 2 = 6(x – 1) y + 2 = 6x – 6 –2 –2 y = 6x – 8 Your Turn: Write an equation in slope-intercept form for the line through the two points. (6, 3) and (0, –1) Step 1 Find the slope. Step 2 Substitute the slope and one of the points into the pointslope form. y – y1 = m(x – x1) Choose (6, 3). Continued Step 3 Write the equation in slope-intercept form. +3 +3 Your Turn: In slope-intercept form, write the equation of the line that contains the points in the table. x –8 –4 y –5 –3.5 4 –0.5 8 1 First, find the slope. Let (x1, y1) be (–8, –5) and (x2, y2) be (8, 1). Next, choose a point, and use either form of the equation of a line. Continued Point-Slope Form Using (8, 1): Rewrite in slope-intercept form. y – y1 = m(x – x1) Distribute. Substitute. Simplify. Solve for y. Standard Form Another way to determine whether a function is linear is to look at its equation. A function is linear if it is described by a linear equation. A linear equation is any equation that can be written in the standard form shown below. Standard Form Notice that when a linear equation is written in standard form. Ax By C •x and y are both on the same side of the equal sign. •x and y both have exponents of 1. •x and y are not multiplied together. • x and y do not appear in denominators, exponents, or radical signs. Most linear equations in Algebra are first written in slope-intercept form: y = mx + b, or point-slope form: y – y1 = m(x – x1). To write a linear equation in standard form you have to transform FROM slope intercept or point slope form TO standard form. Standard form is Ax + By = C. A and B are not BOTH zero. A, B, and C are ALL integers and A is positive. In STANDARD form, the variables are on the left. The constant term is on the right. That just means that you will have to add or subtract to get the variables on the LEFT, the constant on the RIGHT. Be careful if you have coefficients or the constant term that are NOT an integer. You WILL have to get rid of the fraction (by multiplying by the LCD). Example: Write in standard form 2 y x3 5 2 5 y 5( x 3) 5 5 y 2 x 15 5 y 2 x 15 2 x 5 y 15 Original Equation Get rid of the fraction Distribute Subtract 2x from each side Rearrange the terms and multiply by -1 to make A positive. Your Turn: Write in standard form. 3 y 4 ( x 2) 4 3 y 4 ( x 2) 4 Given 4(y + 4) = 3(x – 2) Multiply by 4 to get rid of the fraction. 4y + 16 = 3x – 6) Distributive property 4y = 3x – 22 Subtract 16 from both sides 4y – 3x= – 22 Subtract 3x from both sides – 3x + 4y = – 22 Format x before y 3x – 4y = 22 Multiply by -1 in order to get a positive coefficient for x. Example: Write the standard form. Line passes through (-4,3) with slope -2. y y1 m( x x1 ) Point-slope form y 3 2( x (4)) Substitute y 3 2( x 4) Simplify y 3 2 x 8 Distribute 2 x y 5 Put in standard form Your Turn: Write the standard form of an equation of the line passing through (-6, -3) and slope -1/2. 1 y 3 ( x 6) 2 2(y +3) = -1(x +6) 2y + 6 = -x – 6 2y = -x – 12 2y + x= -12 x + 2y = -12 Given Multiply by 2 to get rid of the fraction. Distributive property Subtract 6 from both sides Add x to both sides Format x before y Your Turn: Write the standard form of an equation of the line passing through (5, 4) and slope -2/3 2 y 4 ( x 5) 3 3(y - 4) = -2(x – 5) Given Multiply by 3 to get rid of the fraction. 3y – 12 = -2x +10 Distributive property 3y = -2x +22 Add 12 to both sides 3y + 2x= 22 Add 2x to both sides 2x + 3y = 22 Format x before y Example: Write the standard form of an equation of the line passing through (5, 4) and (6, 3). y2 y1 m First find slope of the line. x2 x1 m 3 4 1 1 Substitute values and solve for m. 65 1 y 4 1( x 5) y – 4 = -x + 5 y = -x + 9 y+x=9 Put into point-slope form for conversion into Standard Form Ax + By = C Distributive property Add 4 to both sides. Add x to both sides x + y = 9 Standard form requires x come before y. Your Turn: Write the standard form of an equation of the line passing through (-5, 1) and (6, -2) y2 y1 m First find slope of the line. x2 x1 m 2 1 3 3 6 (5) 6 5 11 3 ( x 5) 11 11(y – 1) = -3(x + 5) y 1 11y – 11 = -3x – 15 11y = -3x – 4 11y + 3x = -4 3x + 11y = -4 Substitute values and solve for m. Put into point-slope form for conversion into Standard Form Ax + By = C Multiply by 11 to get rid of fraction Distributive property Add 4 to both sides. Add x to both sides Standard form requires x come before y. Caution When writing a linear equation, NEVER leave it in point-slope form, always transform it to either slope-intercept form or standard form. Linear Equations Essential Question Big Idea: Equivalence • Why are there different forms for the equation of a line? • Depending on what information you know about the line, use the form that is easiest to write an equation of that line. Assignment • Section 2-4 Part 1, Pg 91; #1 – 6 all, 8 – 22 even.