Rate of Change Algebra Problems

Rate of change algebra problems includes problems related to the change in the quantity value divided by the beyond time. In the point of the function, the difference in y- value   is divided by the difference in x- value for the two individual points. Slope is said to be rate of change. This algebra problem includes slope problems.
Important formula:                                              (y2-y1 )
Rate of change, Slope =      --------------
(x2-x1 ).

 

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In the standard form, y=mx+b, here m is slope which is the rate of change of the equation:
Problem with Two Points(rate of Change Algebra Problems)
Ex : 1      find the  slope of the line that contain two points that is (x1,y1)=(2,5) and (x2,y2)=(3,6).
Sol :          Slope = (y2-y1)/(x2-x1)
= (6-5) / (3-2)
= 1/1
Rate of change, slope= 1
Ex : 2 Find the slope of the line that contains two points that is (7, 5) and (4, 3).
Sol :      Here (x1, y1) = (7, 5)
And (x2, y2) = (4, 3)
Slope = (y2-y1)/(x2-x1)
= (3-5) / (4-7)
= (-2)/ (-3)
= 2/3
Rate of change, slope is 2/3
Rate of Change Algebra Problems-problems Related to Parallel and Perepndicular Lines.

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Ex : 3   Find the slope of the line which is parallel to the line that contains two points that is (7, 5) and (4, 4).
Sol :     Here (x1, y1) = (7, 5)
And (x2, y2) = (4, 4)
Slope = (y2-y1)/(x2-x1)
= (4-5) / (4-7)
= (-1)/ (-3)
= 1/3
Slope of the line that contain points is 1/3
Slope of the parallel line is 1/3
Note: slope of the parallel lines are always are same.
Ex : 4  Find the slope of the line which is perpendicular to the line that contains two points that is (7, 5) and (4, 4).
Sol :      Here (x1, y1) = (8, 5)
And (x2, y2) = (4, 4)
Slope = (y2-y1)/(x2-x1)
= (4-5) / (4-8)
= (-1)/ (-4)
= 1/4
Slope of the line that contains points is 1/4
Slope of the perpendicular line is – 1/1/4 = -4
Slope of the perpendicular line is -4
Note: product of slopes of the perpendicular lines (m1.m2) =-1