Question

Draw the graph of x + y=6 which intersects the X axis at A and B respectively. On the same graph paper, draw the graph of 2x-y=3 which intersects the X-axis and Y-axis at C and D respectively. E is the point of intersection of both the graphs. Find the areas of (a) ∆ECA (b) ∆EBD​



ONLY CORRECT ANSWER

Answer 1

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

Given pair of line is

$\rm :\longmapsto\:x + y = 6 - - - (1)$

and

$\rm :\longmapsto\:2x - y = 3 - - - (2)$

Now,

Consider Line (1)

$\rm :\longmapsto\:x + y = 6$

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

$\rm :\longmapsto\:0 + y = 6$

$\rm :\longmapsto\:y = 6$

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

$\rm :\longmapsto\:x + 0 = 6$

$\rm :\longmapsto\:x= 6$

Hᴇɴᴄᴇ,

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

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

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

➢ See the attachment graph.

Now, Consider

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

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

$\rm :\longmapsto\:2(0)- y = 3$

$\rm :\longmapsto\:- y = 3$

$\rm :\longmapsto\:y = - \: 3$

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

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

$\rm :\longmapsto\:2x = 3$

$\rm :\longmapsto\:x = 1.5$

Hᴇɴᴄᴇ,

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

$\begin{gathered}\boxed{\begin{array}{c|c} \bf x & \bf y \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf 0 & \sf - 3 \\ \\ \sf 1.5 & \sf 0 \end{array}} \\ \end{gathered}$

➢ Now draw a graph using the points (0 , - 4) & (1.5 , 0)

➢ See the attachment graph.

Now,

COORDINATES OF A ( 6, 0 )

COORDINATES OF B ( 0, 6 )

COORDINATES OF C ( 1.5, 0 )

COORDINATES OF D ( 0, - 3 )

COORDINATES OF E ( 3, 3 )

So,

$\rm :\longmapsto\:Area_{\triangle ECA} = \dfrac{1}{2} \times 4.5 \times 3 = 6.75 \: sq. \: units$

and

$\rm :\longmapsto\:Area_{\triangle EBD} = \dfrac{1}{2} \times 9 \times 3 = 13.5 \: sq. \: units$