Equation+of+a+line+Period+7

//**Equation of a Line with parallel and perpendicular lines **// //**Carrie, Cori, Emma, Lauren **// #9: Write an equation of a line that is parallel to a given line. Write an equation of a line that is perpendicular to a given line.

1. Write an equation of the line that is parallel to y= 3x-4 and passes through point (0,-3).

Solution to Question 1: y-y1 = m*(x-x1) where (x1,y1) is the given point (0,-3) y -(-3) = 3*(x - 0) y+3 = 3x y = 3x - 3 slope-intercept form 3x-y = 3 standard form

2. Write an equation of a line that passes through the point (2, -2) and is perpendicular to the line y=2x+1. [|Source.] Solution to Question 2: (y+2) = -1/2 (2-x) y + 2 = -1 + 1/2x y=-1/2x + 2

3. Write an equation of the line that passes through the point (-6, -5) and is parallel to line y=5x-1. [|Source.] Solution to Question 3: (y+5) = 5(x+6) y+5 = 5x + 30 y=5x + 25

1.) Find the equation of a line that is parallel to y=4x-2 that passes through the point (6,8).

2.) Find the equation of a line that is perpendicular to y=7x-3 that passes through the point (-14,5).

Solution to question 1 y-y1=m*(x-x1) m stands for the slope x1 and y1 stand for the point (x,y) plug in the same slope as the one in the original equation, because parallel lines have the same slope y-y1=4*(x-x1) Then plug in the new point (6,8) into the equation y-8=4*(x-6) Solve y-8=4x-24 Add 8 to each side y-8+8=4x-24+8 simplify y=4x-32

Solution to question 2 y-y1=m*(x-x1) The m stands for the slope x1 and y1 stand for the new point (x,y) plug in the opposite reciprocal of the original slope from the equation, because perpendicular lines have opposite slopes y-y1=(-1/7)*(x-x1) Then plug in the new point (-14,5) into the equation y-5=(-1/7)*(x-(-14)) Solve y-5= -1/7x-2 Add 5 to each side y-5+5= -1/7x-2+5 simplify y= -1/7x-7

For more help on this subject go to http://www.youtube.com/watch?v=PeL_WMkI7Bw cut and paste this website into your browser. P.S. it is a capital i not a 1 before the 7.