Draw the graph of the pair of linear equations x -2y = 4 and 3x + 5y = 1. Write the vertices of the triangle formed by these two lines and the y-axis. Also find the area of the triangle formed.
Answers
Given : pair of linear equations x -2y = 4 and 3x + 5y = 1. and Y axis
To find : vertices of the triangle formed by these two lines and the y-axis.
Solution:
x -2y = 4 and 3x + 5y = 1 and y axis ( x = 0)
x -2y = 4 and 3x + 5y = 1 intersection
3x - 6y = 12
3x + 5y = 1
=> -11y = 11
=> y = - 1
x = 2
(2 , - 1)
x -2y = 4 and y axis ( x = 0)
=> (0 , - 2)
3x + 5y = 1 and y axis ( x = 0)
=> y = 1/5
(0 , 1/5)
Vertex of Triangle
(0 , 1/5) , (0 , - 2) , (2 , - 1)
Base = (-2 - 1/5) = 11/5
Height = 2
Area = (1/2) (11/5) 2 = 11/5 sq units
or Area = (1/2) | 0 (-2 + 1) + 0(-1 - 1/5) + 2(1/5 -(-2)) | = 11/5 sq units
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Answer:
Hey Your Answer is Here !
Given equations are
x−y−5=0...(1)
3x+5y−15=0...(2)
Write y in terms of x for equation (1).
x−y−5=0
⇒y=(x−5)
Substitute different values of x in the above equation to get corresponding values of y
For x=5,y=0
For x=0,y=−5
For x=1,y=−4
Now plot the points A(5,0), B(0,−5) and C(1,−4) in the graph paper and join A, B and C to get the graph of x−y−5=0
Similarly, Write y in terms of x for equation (2).
3x+5y−15=0
⇒y=
5
(15−3x)
Substitute different values of x in the above equation to get corresponding values of y
For x=5,y=0
For x=0,y=3
For x=−5,y=6
Now plot the points D(5,0), E(0,3) and F(−5,6) in the graph paper and join D, E and F to get the graph of 3x+5y−15=0
From the graph:
Both the lines intersect each other at point A(5,0) and y-axis at B(0,−5) and E(0,3) respectively
∴ Area of △BAE=
2
1
(base×altitude)
=
2
1
×8×5 sq.units
=20 sq.units.