Draw the graphs of x – y + 1 = 0 and 3x + 2y – 12 = 0. Determine the coordinates of the vertices of the triangle formed by these lines and x-axis and shade the triangular area. Calculate the area bounded by these lines and x-axis.
Answers
The graphs of x – y + 1 = 0 and 3x + 2y – 12 = 0 are as shown in the below figure.
The area bounded by these lines and y-axis is 15/2 sq.units.
Given equations are:
x – y + 1 = 0 and 3x + 2y – 12 = 0
x - y + 1 = 0
when x = 0, y = -1
when y = 0, x = 1
We have obtained 2 points (0, -1) and (1, 0)
3x + 2y – 12 = 0
when x = 0, y = 6
when y = 0, x = 4
We have obtained 2 points (0, 6) and (4, 0)
The coordinates of the vertices of the triangle formed by these lines and x-axis are:
(2, 3), (-1, 0) and (4, 0)
The area bounded by these lines and x-axis is:
= 1/2 [x1 ( y2 - y3 ) + x2 ( y3 - y1 ) + x3 ( y1 - y2 )]
= 1/2 [2 ( 0 - 0 ) + (-1) (0 - 3 ) + 4 ( 3 - 0 )]
= 1/2 [ 2 (0) - 1 (-3) + 4 (3) ]
= 1/2 [ 0 + 3 + 12 ]
= 1/2 [ 15 ]
= 15/2 sq. units.
Answer:
The graphs of x – y + 1 = 0 and 3x + 2y – 12 = 0 are as shown in the below figure.
The area bounded by these lines and y-axis is 15/2 sq.units.
Given equations are:
x – y + 1 = 0 and 3x + 2y – 12 = 0
x - y + 1 = 0
when x = 0, y = -1
when y = 0, x = 1
We have obtained 2 points (0, -1) and (1, 0)
3x + 2y – 12 = 0
when x = 0, y = 6
when y = 0, x = 4
We have obtained 2 points (0, 6) and (4, 0)
The coordinates of the vertices of the triangle formed by these lines and x-axis are:
(2, 3), (-1, 0) and (4, 0)
The area bounded by these lines and x-axis is:
= 1/2 [x1 ( y2 - y3 ) + x2 ( y3 - y1 ) + x3 ( y1 - y2 )]
= 1/2 [2 ( 0 - 0 ) + (-1) (0 - 3 ) + 4 ( 3 - 0 )]
= 1/2 [ 2 (0) - 1 (-3) + 4 (3) ]
= 1/2 [ 0 + 3 + 12 ]
= 1/2 [ 15 ]
= 15/2 sq. units.