Question

plot a graph for each of the following pairs of equation and shade the region bounded by the 2 lines and the x axis i) x-y+1 =0 and 2x+y-10=0​

Answer 1

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

Given pair of linear equations are

$\rm :\longmapsto\:x - y + 1 = 0 - - - (1)$

and

$\rm :\longmapsto\:2x + y - 10 = 0 - - - (2)$

Now,

Consider, Equation (1),

$\rm :\longmapsto\:x - y + 1 = 0$

can be rewritten as

$\rm :\longmapsto\:x - y = - \: 1$

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

$\rm :\longmapsto\:0 - y = - \: 1$

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

$\rm :\longmapsto\: y = \: 1$

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

$\rm :\longmapsto\:x - 0 = - \: 1$

$\rm :\longmapsto\:x = - \: 1$

Hᴇɴᴄᴇ,

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

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

➢ Now draw a graph using the points (0 , 1) & (- 1 , 0)

➢ See the attachment graph. [ Black Line ]

Consider Equation (2),

$\rm :\longmapsto\:2x + y - 10 = 0$

can be rewritten as

$\rm :\longmapsto\:2x + y = 10$

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

$\rm :\longmapsto\:2x + 0 = 10$

$\rm :\longmapsto\:2x = 10$

$\rm :\longmapsto\:x = 5$

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

$\rm :\longmapsto\:2 \times 0 + y = 10$

$\rm :\longmapsto\:0 + y = 10$

$\rm :\longmapsto\:y = 10$

Hᴇɴᴄᴇ,

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

$\begin{gathered}\boxed{\begin{array}{c|c} \bf x & \bf y \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf 0 & \sf 10 \\ \\ \sf 5 & \sf 0 \end{array}} \\ \end{gathered}$

➢ Now draw a graph using the points (0 , 10) & (5 , 0)

➢ See the attachment graph. [ Red Line ]

Hence, the required region is triangle ABC.