Question

Draw the graphs of the following linear equations on the same graph paper:
2x + 3y = 12, x − y = 1

Find the coordinates of the vertices of the triangle formed by the two straight lines and the area bounded by these lines and x-axis.

Answer 1

We are given,

We get,

Now, substituting in , we get

Substituting in , we get

Thus, we have the following table exhibiting the abscissa and ordinates of points on the line represented by the given equation

x	0	6
y	4	0

Plotting A(0,4) and E(6,0) on the graph and by joining the points , we obtain the graph of equation .

We are given,

We get,

$y=x-1$

Now, substituting in $y=x-1$,we get

$y=-1$

Substituting in $y=x-1$,we get

$y=-2$

Thus, we have the following table exhibiting the abscissa and ordinates of points on the line represented by the given equation

x	0	–1
y	-1	-2

Plotting D(0,1) and E(-1,0) on the graph and by joining the points , we obtain the graph of equation .

By the intersection of lines formed by and on the graph, triangle ABC is formed on y axis.

Therefore,

AC at y axis is the base of triangle ABC having AC = 5 units on y axis.

Draw FE perpendicular from F on y axis.

FE parallel to x axis is the height of triangle ABC having FE = 3 units on x axis.

Therefore,

Area of triangle ABC, say A is given by

$$\begin{gathered} A=\frac{1}{2}\left(\mathrm{Base}\times \mathrm{Height}\right) \\ =\frac{1}{2}\left(\mathrm{AC}\times \mathrm{FE}\right) \\ =\frac{1}{2}\left(5\times 3\right) \\ =\frac{15}{2}\mathrm{sq}.\mathrm{units} \end{gathered}$$