LINEAR EQUATIONS YEA!. WHAT IS A LINEAR EQUATION? A straight line Have simple variable equations No exponents Only have x and y 3 different equations.

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LINEAR EQUATIONS YEA!

WHAT IS A LINEAR EQUATION? A straight line Have simple variable equations No exponents Only have x and y 3 different equations

SLOPE-INTERCEPT FORM Equation is y=mx+b Called this because m is the slope and b is the y-intercept

EXAMPLE 1 (FINDING B) Find equation with slope of m=4 and passes through the points (-1,-6) y=mx+b (-6)=(4)(-1)+b -6=-4+b -2=b Full equation : y=4x-2

EXAMPLE 2 (FINDING SLOPE) Find equation that passes through the points (-2,4) and (1,2) Use slope formula which is: m = [y1 - y2] / [x1 - x2] (4-2)/(-2-1)=-2/3 Then choose any of the two points ( doesn’t matter) and plug into slop intercept formula y=mx+b 4=-2/3(-2) +b 4=4/3+b 4-4/3=b 12/3-4/3=b b=8/3 Final equation : y=(-2/3)x+8/3

POINT-SLOPE FORM Equation is y-y1=m(x-x1) You are given a slope and a point when using this formula

EXAMPLE 1 (SIMPLE) Find equation of straight line that has a slope of m=4 and passes through (-1,-6) So m=4, y1=-6, x1=-1 y-y1=m(x-x1) y-(-6)=(4)(x-(-1)) y+6=4(x+1) y+6=4x+4 y=4x+4-6 Final equation: y=4x-2

EXAMPLE 2 (LOWKEY COMPLEX) Find the equation of the line that passes through the points (-2,4) and (1,2) First, find slope using m = [y1 - y2] / [x1 - x2]= (4-2)/(-2-1)=-2?3 Then use point-slope form y-y1=m(x-x1) y-4=(-2/3)(x-(-2)) y=(-2/3)x-4/3+4 y=(-2/3)x-4/3+12/3 y=(-2/3)x+8/3

FINDING PARALLEL AND PERPENDICULAR LINES (YAY…) Fairly easy All you do is for parallel use identical slope For perpendicular use opposite slope (REVERSE IT)

EXAMPLE 1 (PARALLEL) Given the line 2x-3y=9 and the point (4,-1), find the parallel line Solve 2x-3y=9 -3y=-2x+9 y=(2/3)x-3 reference slope is: m=(2/3) Plug in points to y-y1=m(x-x1) You get y=(2/3)x-11/3 Parallel line: y=(2/3)x-11/3

EXAMPLE 2 (PERPENDICULAR) Flip the reference slope to (-3/2) Plug into y-y1=m(x-x1) using (4,-1) Perpendicular line: y=(-3/2)x+5