Linear Equations ReviewWhat you should know about Linear equationsEquation FormsSlopesCan you run through the linear equation information…Given any linear equation, one should be able to 3x+4y=24Given any linear equation, one should be able to y = 1/2x-7Given any linear equation, one should be able to y = 5Given any linear equation, one should be able to x = 6To graph a lineGraph using intercepts y = -3x – 7Slide 12The slope formulaFind the slope given 2 points (-1,1); (2,3)Now write the equation (-1,1); (2,3)If you have two points you can find the line…sometimes the challenge is knowing what you have.Parallel & Perpendicular II |Linear equation partsGiven 3x + 4y = 12Slide 20Slide 21Given y = 2x - 12Slide 23The alternative calculation is to using the point slope form of a linear equation y – y1 = m(x – x1)Find (e,f)remember if you can find the blue lineFind tSlide 28Switching gears…Linear Equations ReviewLinear Equations ReviewChapter 5Chapters 1 & 2What you should know aboutWhat you should know aboutLinear equationsLinear equationsSlopeY-interceptX-interceptWhat does the graph look like?Parallel slopePerpendicular slopeGiven ANY linear equation you should be able to identify…Equation FormsEquation FormsSlope InterceptStandardHorizontalVerticaly = mx + bAx + By = Cy = bx = aSlopesSlopesNegativePositiveHorizontal VerticalCan you run through the Can you run through the linear equation information…linear equation information… 3x+4y=243x+4y=24 y = 1/2x-7y = 1/2x-7 y = 5y = 5 x = 6x = 6Given any linear equation, one Given any linear equation, one should be able to should be able to 3x+4y=243x+4y=24The Equation FormDirectionSlopey-interceptx-interceptParallel SlopePerpendicular Slope1. Standard2. Falling3. -3/4 4. 65. 86. -3/4 7. 4/3identify…Given any linear equation, one Given any linear equation, one should be able to should be able to y = 1/2x-7y = 1/2x-7The Equation FormDirectionSlopey-interceptx-interceptParallel SlopePerpendicular Slope1. Slope intercept2. Rising3. 1/24. -75. - -7/(1/2) = 146. 1/27. -2identify…Given any linear equation, one Given any linear equation, one should be able to should be able to y = 5y = 5The Equation FormDirectionSlopey-interceptx-interceptParallel SlopePerpendicular Slope1. Horizontal line2. horizontal3. 04. 55. Does not exist6. 07. undefinedidentify…Given any linear equation, one Given any linear equation, one should be able to should be able to x x = 6 = 6The Equation FormDirectionSlopey-interceptx-interceptParallel SlopePerpendicular Slope1. Vertical line2. verticle3. undefined 4. Does not exist5. 66. undefined 7. 0identify…To graph a lineTo graph a lineIntercepts◦Identify the intercepts◦Plot the intercepts◦Draw the linePoint-slope◦Identify a point on the line and the slope◦Plot the point◦Count the slope “Rise/Run”(0,-7)(-7/3,0)Graph using intercepts Graph using intercepts y = y = -3x – 7-3x – 7y-int = -7x-int = 7 -3up 3 back 1(0,-7)down 3 over 1Graph using intercepts Graph using intercepts y = y = -3x – 7-3x – 7point= (0, -7)slope = -3 / 1The slope formulaThe slope formulam = y1 – y2 x1 – x2This is really the same at the point-slope equationm(x1 – x2) = y1 – y2y1 – y2 = m(x1 – x2)Find the slope given 2 points (-Find the slope given 2 points (-1,1); (2,3)1,1); (2,3)m = 3 – 1 2 – -1m = 2 3m = 1 – 3 -1 – 2m = -2 = 2 -3 3Now write the equation (-1,1); Now write the equation (-1,1); (2,3)(2,3)m = 3 – 1 2 – -1m = 2 3y1 – y2 = m(x1 – x2)y – 3 = 2/3 (x – 2)y = 2/3 x – 4/3 + 9/3y = 2/3 x + 5/3If you have two points you can find If you have two points you can find the line…sometimes the challenge is the line…sometimes the challenge is knowing what you have.knowing what you have.Given◦The origin◦The y-intercept◦The x-intercept◦A line parallel to the x-axis◦A line parallel to the y-axisYou have …◦the point (0,0)◦the point (0,y)◦the point (x,0)◦the slope m = 0 eqn is y = ____◦The slope m undefined eqn is x = ____Parallel & PerpendicularParallel & Perpendicular II II | | Parallel slopes are equalm originalm|| = moPerpendicular slopes are opposite reciprocalsm originalm | = -1 / moLinear equation partsLinear equation partsSlope InterceptStandard Horizontal VerticalEquationy = mx + bAx + By = C y = b x = aSlope m-A B0 undefinedy – interceptbC Bbdoes not existx - intercept-b mC Adoes not existaparallel slope || m-A B0 undefinedperpendicular slope _|_ -1 mB Aundefined 0Given 3x + 4y = 12Given 3x + 4y = 12Find the line || to the given line that passes through (-2, 5).Find the line _|_ to the given line that passes through (-2, 5).Given 3x + 4y = 12Given 3x + 4y = 12Find the line || to the given line that passes through (-2, 5).Find the line _|_ to the given line that passes through (-2, 5).If the line is parallel then the slope must be the same so the linear equation will look like 3x + 4y = If the line is perpendicular then the slope must be the opposite reciprocal so the linear equation will look like -4x + 3y = Given 3x + 4y = 12Given 3x + 4y = 12Find the line || to the given line that passes through (-2, 5).3x + 4y = ___3(-2) + 4(5) = ___-6 + 20 = 14Find the line _|_ to the given line that passes through (-2, 5).-4x + 3y = ___-4(-2)+3(5) = ___8+15 = 233x + 4y = 143x + 4y = 14-4x + 3y = 23-4x + 3y = 23Given y = 2x - 12Given y = 2x - 12Find the line || to the given line that passes through (-2, 5).Slope = 2 thereforem|| = 2Find the line _|_ to the given line that passes through (-2, 5).Slope = 2 thereforem_|_ = -1/2 y = 2x + by = 2x + by = -1/2 x + by = -1/2 x + bGiven y = 2x - 12Given y = 2x - 12Find the line || to the given line that passes through (-2, 5).Find the line _|_ to the given line that passes through (-2, 5).y = 2x + by = 2x + b5 = 2(-2) + b5 = 2(-2) + bb = 9b = 9y = -1/2 x + by = -1/2 x + b5 = -1/2 (-2) + 5 = -1/2 (-2) + bbb = 4b = 4y = -1/2 x + 4y = -1/2 x + 4y = 2x + 9y = 2x + 9The alternative calculation is The alternative calculation is to using the point