One of the most important and exam oriented question as it was already asked in previous year paper of 2009 in which we have given the equations x-y-5=0, 3x+5y-15=0 and 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

RS Aggarwal, Class 10, chapter 3A, question no 18

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=(15−3x)/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=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=21(base×altitude)

=1/2×8×5 sq.units

=20 sq.units.