Draw the graphs of the following system of linear
equations 3x + y = 3; 3x + y = 9. Find the
coordinates of the vertices of the trapezium
formed by x-axis, y-axis and these lines. Also,
find the area of the trapezium so formed. pleaseanswer dedo
Answers
Refer the attachment for the graph of linear equations
3x + y = 3
3x + y = 9.
A) The coordinates of the vertices of trapezium for the equation 3x + y = 3 formed by
x - axis is (1,0)
y - axis is (0,3)
B) The coordinates of the vertices of trapezium for the equation 3x + y = 9 formed by
x - axis is (3,0)
y - axis is (0,9)
C) The area of trapezium is 12 sq.units.
=>
A) For the equation 3x + y = 3
- x - axis
assume y value as 0
= 3x + (0) = 3
= 3x = 3
= x = 1
Coordinates of x axis is (1,0) for the equation 3x + y = 3.
- y - axis
assume x value as 0
= 3(0) + y = 3
= y = 3
Coordinates of y axis is (0,3) for the equation 3x + y = 3.
B) For the equation 3x + y = 9
- y - axis
assume x value as 0
= 3(0) + y = 9
= y = 9
Coordinates of y axis is (0,9) for the equation 3x + y = 9.
- x - axis
assume y value as 0
= 3x + (0)
= 3x = 9
= x = 3
Coordinates of x axis is (3,0) for the equation 3x + y = 9.
C) The area of trapezium
To find the area of trapezium the formula is
Area of trapezium = area of triangle 1 - area of triangle 2
and to find the area of triangle
area of traingle = 1/2 * base * height.
Assume ABCXY as trapezium in which
AXYC is a trapezium
ABC is a triangle (1)
XBY is a triangle (2)
then area of triangle (1) ABC
= 1/2 * 3 * 9
= 27/2 sq.units.
area of triangle (2) XBY
= 1/2 * 1 * 3
= 3/2 sq.units.
Hence
the area of trapezium(AXYC) is
= 24/2 - 3/2
= 12 sq.units.
