Question

draw the graph representing the equation x - y= 1 and 2 X + 3 Y = 12 on the same graph paper find the area of a triangle formed by these lines at x-axis and the y-axis​

Answer 1

$\large\underline{\sf{Solution-}}$

Consider the linear equation

$\rm \: x - y = 1$

Substituting 'x = 0' in the given equation, we get

$\rm \: 0 - y = 1$

$\rm \: - y = 1$

$\rm\implies \:y = - 1$

Substituting 'y = 0' in the given equation, we get

$\rm \: x - 0 = 1$

$\rm\implies \:x = 1$

Hᴇɴᴄᴇ,

➢ Pair of points of the given equation are shown in the below table.

$\begin{gathered}\boxed{\begin{array}{c|c} \bf x & \bf y \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf 0 & \sf - 1 \\ \\ \sf 1 & \sf 1 \end{array}} \\ \end{gathered}$

Now, Consider the linear equation

$\rm \: 2x + 3y = 12$

Substituting 'x = 0' in the given equation, we get

$\rm \: 2 \times 0 + 3y = 12$

$\rm \: 0 + 3y = 12$

$\rm \: 3y = 12$

$\rm\implies \:y = 4$

Substituting 'y = 0' in the given equation, we get

$\rm \: 2x + 3 \times 0 = 12$

$\rm \: 2x + 0 = 12$

$\rm \: 2x = 12$

$\rm\implies \:x = 6$

Hᴇɴᴄᴇ,

➢ Pair of points of the given equation are shown in the below table.

$\begin{gathered}\boxed{\begin{array}{c|c} \bf x & \bf y \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf 0 & \sf 4 \\ \\ \sf 6 & \sf 0 \end{array}} \\ \end{gathered}$

➢ Now draw the lines using the points (0 , 4), (6 , 0), (1, 0) & (0 , - 1)

➢ See the attachment graph.

Now, Area of triangle bounded by lines with x - axis

$\color{green}\boxed{ \rm{ \:Area_{(\triangle)} = \frac{1}{2} \times 5 \times 2 = 5 \: sq. \: units \: }}$

Now, Area of triangle bounded by lines with y - axis

$\color{blue}\boxed{ \rm{ \:Area_{(\triangle)} = \frac{1}{2} \times 5 \times 3 = 7.5 \: sq. \: units \: }}$

Answer 2

Answer:

7.5 sq.units

Step-by-step explanation:

Graph of the equation 2x + 3y - 12 = 0

we have,

$2x = 12 - 3y$  

→ `$x=\frac{12- 3y }{2}$

Putting y = 4, we get `x ( 12 - 3 x  4)/2 = 0

Putting y = 2, we get `x = (12 - 3 x 2 )/2= 0

Thus, we have the following table for the p table for the points on the line 2x + 3y = 12

Plotting points A ( 0,4 ) , B (3,2) on the graph paper and drawing a line passing through them

we obtain graph of the equation.

Graph of the equation

Graph of the equation x-y - 1 :

We have x   y = 1 = x = 1  y

Thus, we have the following table for the points the line x - y = 1

Plotting points C(1,0) and D (0,-1) on the same graph paper drawing a line passing through  the m, we obtain the graph of the line represents by the equation x - y = 1

Clearly two lines intersect at A (3, 2)

The graph of time 2x + 3y = 12 = intersect with y-axis at B (0,4) and the graph of the line

X - y = 1 intersect with y axis at c ( =  0,-1).

So, the vertices of the triangle formed by thee two straight lines and y-axis are A (3,2) and

B( 0.4) and C (0,-1)

Now,

Area of `AABC = 1/2` [ Base x (x) = Height]

= 1/2 ( Bc x x AB)` =

= 1/2 ( 5 + 3)`

= 15/2

= 7.5 sq.units