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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
2xâˆ’3y+6=0...(1)
2x+3yâˆ’18=0......(2)

WriteÂ yÂ in terms ofÂ xÂ for equationÂ (1).
2xâˆ’3y+6=0
â‡’y=(2x+6)/3â€‹
Substitute different values ofÂ xÂ in the above equation to get corresponding values ofÂ y

ForÂ Â x=âˆ’3,y=0
ForÂ Â x=0,y=2
ForÂ Â x=3,y=4

Now plot the pointsÂ A(âˆ’3,0),Â B(0,2)Â andÂ C(3,4)Â in the graph paper and joinÂ A,Â BÂ andÂ CÂ to get the graph ofÂ 2xâˆ’3y+6=0

Similarly, WriteÂ yÂ in terms ofÂ xÂ for equationÂ (2).
2x+3yâˆ’18=0
â‡’y=(18âˆ’2x)/3â€‹
Substitute different values ofÂ xÂ in the above equation to get corresponding values ofÂ y

ForÂ Â x=0,y=6
ForÂ Â x=3,y=4
ForÂ Â x=6,y=2

Now plot the pointsÂ D(0,6),Â E(3,4)Â andÂ F(6,2)Â in the graph paper and joinÂ D,Â EÂ andÂ FÂ to get the graph ofÂ 2x+3yâˆ’18=0

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

âˆ´Â Area ofÂ â–³CBD=1/2â€‹(baseÃ—altitude)
=1/2â€‹Ã—4Ã—3Â sq.units
=6Â sq.units.

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