State The Equation of Parallel and Perpendicular Lines Harry Cai & Group 5.

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Presentation transcript:

State The Equation of Parallel and Perpendicular Lines Harry Cai & Group 5

Examples of Parallel lines  The following picture shows 2 parallel lines.  From looking at the picture, we can tell that the slope of the parallels are the same

Examples of Perpendicular Lines  Here we have a line perpendicular to another line.  Obliviously, the slope of the perpendicular line is not the same as the other line.

Here are a few skills you should know in order to continue  The Point-Slope Form:  y-y1=m(x-x1)  How to find the slope for a parallel line  How to state a equation of a line when given the slope and one point

To state an equation of a line parallel to a given equation and point  Step 1:Find the slope of the equation  Step 2:Find the slope of the parallel line  Step 3:Sub in know values into a point-slope form  Step 4:Simplify

Example  State an equation of the line parallel to 3x+y-4=0 and through the point A(2, -5)  Step 1: To find the slope, write the equation in y=mx+b form.  3x+y-4=0, y=-3x+4  Therefore, the slope of the line is -3

Example  Step 2: The slope of a line parallel to it is also -3  Step 3:Now we sub in the values  y-y1=m(x-x1)  y-(-5)=-3(x-2)  Step 4: Simplify  y+5=-3x+6  3x+y-1=0  Therefore, the equation of a line parallel is 3x+y-1=0

To state an equation of a line perpendicular to a given equation and point  Step 1:Find the slope of the equation  Step 2:Find the slope of the perpendicular line  Step 3:Sub in the know values into a point-slope form  Step 4:Simplify

Example  State an equation of the line perpendicular to the line 4x+2y-7=0 and through the point A(6,0)  Step 1: To find the slope, write the equation in y=mx+b form  4x+2y-7=0, 2y=-4x+7, y=-2x+7  Therefore, the slope of the line is -2

Example  Step 2: To find the slope of the perpendicular line, we must find the negative reciprocal of the slope of the given equation  Since m=-2, the negative reciprocal of that would be ½  Step 3: Now we sub in the values  y-y1=m(x-x1)  y-0=1/2(x-6)

Example  Step 4: Simplify  y-0=1/2(x-6)  2y=1(x-6)  2y=x-6  -x+2y+6=0  X-2y-6=0  Therefore, the equation of a perpendicular line is x-2y-6=0