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# Solve the equations graphically and find the vertices and the area of the triangle formed by these lines and the y-axis: 2x-3y+6=0, 2x+3y-18=0

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This is the basic and conceptual question from linear equations in two variables in which we have been asked to solve the equations graphically and also asked to find the vertices and the area of the triangle formed by these lines and the y-axis: 2x-3y+6=0, 2x+3y-18=0

RS Aggarwal, Class 10, chapter 3A, question no 16

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1. Given equations are
2x3y+6=0...(1)
2x+3y18=0......(2)

Write y in terms of x for equation (1).
2x3y+6=0
y=(2x+6)/3
Substitute different values of x in the above equation to get corresponding values of y

For  x=3,y=0
For  x=0,y=2
For  x=3,y=4

Now plot the points A(3,0), B(0,2) and C(3,4) in the graph paper and join A, B and C to get the graph of 2x3y+6=0

Similarly, Write y in terms of x for equation (2).
2x+3y18=0
y=(182x)/3
Substitute different values of x in the above equation to get corresponding values of y

For  x=0,y=6
For  x=3,y=4
For  x=6,y=2

Now plot the points D(0,6), E(3,4) and F(6,2) in the graph paper and join D, E and F to get the graph of 2x+3y18=0

From the graph:
Both the lines intersect each other at point C(3,4) and y-axis at B(0,2) and D(0,6) respectively

Area of CBD=1/2(base×altitude)
=1/2×4×3 sq.units
=6 sq.units.

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