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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: 4x-y-4=0, 3x+2y-14=0

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This is the basic and conceptual question from linear equations in two variables in which we have 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 and the given equations are 4x-y-4=0, 3x+2y-14=0

RS Aggarwal, Class 10, chapter 3A, question no 17

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1. Given equations are
4xâˆ’yâˆ’4=0...(1)
3x+2yâˆ’14=0...(2)

WriteÂ yÂ in terms ofÂ xÂ for equationÂ (1).
4xâˆ’yâˆ’4=0
â‡’y=(4xâˆ’4)
Substitute different values ofÂ xÂ in the above equation to get corresponding values ofÂ y

ForÂ Â x=1,y=0
ForÂ Â x=0,y=âˆ’4
ForÂ Â x=2,y=4

Now plot the pointsÂ A(1,0),Â B(0,âˆ’4)Â andÂ C(2,4)Â in the graph paper and joinÂ A,Â BÂ andÂ CÂ to get the graph ofÂ 4xâˆ’yâˆ’4=0

Similarly, WriteÂ yÂ in terms ofÂ xÂ for equationÂ (2).
3x+2yâˆ’14=0
â‡’y=(14âˆ’3x)/2â€‹
Substitute different values ofÂ xÂ in the above equation to get corresponding values ofÂ y

ForÂ Â x=0,y=7
ForÂ Â x=2,y=4
ForÂ Â x=4,y=1

Now plot the pointsÂ D(0,7),Â E(2,4)Â andÂ F(4,1)Â in the graph paper and joinÂ D,Â EÂ andÂ FÂ to get the graph ofÂ 3x+2yâˆ’14=0

From the graph:
Both the lines intersect each other at pointÂ C(2,4)Â andÂ y-axis atÂ B(0,âˆ’4)Â andÂ D(0,7)Â respectively

âˆ´Â Area ofÂ â–³BCD=1/2â€‹(baseÃ—altitude)
=1/2Ã—11Ã—32Â sq.units
=11Â sq.units.

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