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
Given equations are
4x−y−4=0...(1)
3x+2y−14=0...(2)
Write y in terms of x for equation (1).
4x−y−4=0
⇒y=(4x−4)
Substitute different values of x in the above equation to get corresponding values of y
For x=1,y=0
For x=0,y=−4
For x=2,y=4
Now plot the points A(1,0), B(0,−4) and C(2,4) in the graph paper and join A, B and C to get the graph of 4x−y−4=0
Similarly, Write y in terms of x for equation (2).
3x+2y−14=0
⇒y=(14−3x)/2
Substitute different values of x in the above equation to get corresponding values of y
For x=0,y=7
For x=2,y=4
For x=4,y=1
Now plot the points D(0,7), E(2,4) and F(4,1) in the graph paper and join D, E and F to get the graph of 3x+2y−14=0
From the graph:
Both the lines intersect each other at point C(2,4) and y-axis at B(0,−4) and D(0,7) respectively
∴ Area of △BCD=1/2(base×altitude)
=1/2×11×32 sq.units
=11 sq.units.