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The provided answer solutions compute each slope m from the two points, determine the intercept c using substitution, and rewrite the results as equivalent standard-form expressions for six 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Give your answer in the form y = mx + c  \n• Rewrite your answer in standard form Ax + By + C = 0  \n1) Points (-2, 15) and (4, -9) 2) Points (-2, -20) and (5, 1)  \n3) Points (3, -1) and (11, 3) 4) Points (-2, 5) and (10, -13)  \n5) Points (-6, 1) and (9, 11) 6) Points (-3, -6) and (2, 3)  \nName Date  \nLINEAR EQUATION FROM TWO POINTS SHEET 3 ANSWERS  \n• Find the equation on the line passing through the two points.  \n• Give your answer in the form y = mx + c  \n• Rewrite your answer in standard form Ax + By + C = 0  \n1) Points (-2, 15) and (4, -9) 2) Points (-2, -20) and (5, 1)  \nm = (-9 – 15)/(4 –(-2)) = -24/6 = -4 y = mx + c so this means y = -4x + c Use the point (-2, 15)  \nSubstitute values 15 = -4(-2) + cc + 8 = 15 so c = 7  \nSo y = -4x + 7 (or y = 7 – 4x) Standard Form:  \ny = 7 – 4x 4x + y = 7  \nSo 4x + y – 7 = 0  \nm = (1 –(-20))/(5 –(-2)) = 21/7 = 3 y = mx + c so this means y = 3x + c Use the point (-2, -20) Substitute values-20 = 3(-2) + cc + (-6) = -20 so c = -14  \nSo y = 3x-14  \nStandard Form: y = 3x-14  \n3x – 14-y = 0  \nSo 3x – y – 14 = 0  \n3) Points (3, -1) and (11, 3) 4) Points (-2, 5) and (10, -13)  \nm = (3 –(-1)) / (11 – 3) = 4/8 = ½ y = mx + c so this means y = ½ x + c Use the point (3, -1)  \nSubstitute values: -1 = ½ (3) + c So c + 3/2 = -1  \nc = -1 – 3/2 = -5/2 or-2 ½ So y = ½ x – 5/2  \nStandard Formy = ½ x – 5/2  \n½ x – 5/2 – y = 0  \n½ x – y – 5/2 = 0  \nx – 2y – 5 = 0  \nm = (-13 – 5)/(10 –(-2)) = -18/12 = -3/2  \ny = mx + c so y = -3/2 x + c Use the point (-2, 5)  \nSubstitute values: 5 = -3/2 (-2) + cc + 3 = 5 so c = 2  \nSo y = -3/2 x + 2 or 2 – 3/2 x  \nStandard Formy = 2 – 3/2 xy + 3/2 x – 2 = 0  \n2y + 3x – 4 = 0  \n3x + 2y – 4 = 0  \n5) Points (-6, 1) and (9, 11) 6) Points (-3, -6) and (2, 3)  \nm = (11 – 1)/(9 –(-6)) = 10/15 = 2/3  \ny = mx + c so y = 2/3 x + c Use the point (-6, 1)  \nSubstitute values: 1 = 2/3 (-6) + cc – 12/3 = 1  \nc – 4 = 1 so c = 5  \nSo y = 2/3 x + 5 Standard Formy = 2/3 x + 5 2/3 x + 5 – y = 0  \n2/3 x – y + 5 = 0  \n2x – 3y + 15 = 0  \nm = (3 –(-6)) / (2 –(-3)) = 9/5 y = mx + c so y = 9/5 x + c Use the point (-3, -6) Substitute values: -6 = 9/5 (-3) + cc – 27/5 = -6  \nc = -6 + 27/5 = -30/5 + 27/5 = -3/5 So y = 9/5 x – 3/5  \nStandard Formy = 9/5 x – 3/5  \n9/5 x – 3/5 – y = 0  \n9/5 x – y – 3/5 = 0  \n9x – 5y – 3 = 0","cbCait8YBX5wsOWu","https://ap.wps.com/l/cbCait8YBX5wsOWu","pdf",197310,"English","# LINEAR EQUATION FROM TWO POINTS SHEET 3\n# LINEAR EQUATION FROM TWO POINTS SHEET 3 ANSWERS","[{\"question\":\"What steps are used to form the equation of a line from two points?\",\"answer\":\"Find the slope m using the two coordinates, write y = mx + c, substitute one point to solve for c, then rewrite the equation in standard form Ax + By + C = 0.\"},{\"question\":\"How is the slope m calculated in these problems?\",\"answer\":\"Use the change in y divided by the change in x between the two points, for example m = (y2 - y1) / (x2 - x1).\"},{\"question\":\"What is the required final format after finding y = mx + c?\",\"answer\":\"Rewrite the equation as standard form Ax + By + C = 0, showing equivalent expressions for each pair of points.\"}]","LINEAR EQUATION FROM TWO POINTS SHEET 3 | PDF",1792677586]