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