Question

Draw the graph of following pair of linear equations in two variables

x + y = 4

3x – 2y =12

Answer the questions below

(i) Solution of the given equations.

(ii) Vertices of the triangle formed by given lines and y-axis.

(iii) Area of the triangle formed.

(iv) Co-ordinates of the points where lines cut the y-axis.​

Answer 1

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

Given pair of linear equations are

$\rm :\longmapsto\:x + y = 4$

and

$\rm :\longmapsto\:3x - 2y = 12$

Consider,

$\rm :\longmapsto\:x + y = 4$

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

$\rm :\longmapsto\:0 + y = 4$

$\bf\implies \:y = 4$

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

$\rm :\longmapsto\:x + 0 = 4$

$\bf\implies \:x = 4$

Hᴇɴᴄᴇ,

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

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

➢ Now draw a graph using the points (0 , 4) & (4 , 0)

➢ See the attachment graph. [ Purple line ]

Consider,

$\rm :\longmapsto\:3x - 2y = 12$

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

$\rm :\longmapsto\:3 \times 0 - 2y = 12$

$\rm :\longmapsto\: - 2y = 12$

$\bf\implies \:y = - 6$

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

$\rm :\longmapsto\:3x - 2 \times 0 = 12$

$\rm :\longmapsto\:3x - 0 = 12$

$\rm :\longmapsto\:3x = 12$

$\bf\implies \:x = 4$

Hᴇɴᴄᴇ,

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

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

➢ Now draw a graph using the points (0 , - 6) & (4 , 0)

➢ See the attachment graph. [ Blue line ]

From graph, we concluded that

Solution:- i

The solution is x = 4 and y = 0.

Solution :- ii

The vertex of triangle thus formed are (4, 0), (0, - 6) and (0, 4).

Solution :- iii

$\rm :\longmapsto\:Area_{\triangle} = \dfrac{1}{2} \times 4 \times 10 = 20 \: sq. \: units$

Solution :- iv

Line (1) cut the y - axis at (0, 4)

Line (2) cut the y - axis at (0, - 6).