1 Answer | Add Yours
Now here we have a equation of a straight line. And also we have to lines on the x-axis. We need to find the area which is formed by these lines.
Now let us see the shape of these lines.
y = 1/2x+3
This is a inclined line which is forming a positive angle with x-axis.
These lines are vertical straight lines which goes through points (-1,0) and (5,0) respectively.
Now if we consider the shape of the geometry we obtain from these lines, it is a trapizoid.
We already know the base line and it's points in the trapizoid. They are (-1,0) and (5,0)
Now let's find the points which is on the inclined leg of the trapizoid.
These points are can be obtained by solving y = 1/2x+3 at x=-1 and x=5
y = 1/2x+3 when x=-1 then y = 1/2(-1)+3 = 5/2
y = 1/2x+3 when x=5 then y = 1/2(5)+3 = 11/2
So the points are (-1,5/2),(-1,0),(5,0),(5,11/2)
Let us name these points A,B,C,D respectively.
From these points;
Line AB form by (-1,0) and (-1,5/2) is parallel to line CD form by (5,0) and (5,11/2).
The base or the perpendicular line (BC) is formed by (-1,0) and (5,0).
So area of the trapizoid = (AB+CD)*BC/2
AB = `sqrt((-1-(-1))^2+(5/2-0)^2)` = 5/2
BC = 6
CD = 11/2
So area of the trapizoid = (5/2+11/2)*6/2 = 24
Area under the graph of y= ½ x+3 between x=-1 and x=5 is 24.
The following graph shows it's sketch.
We’ve answered 319,852 questions. We can answer yours, too.Ask a question