Question

draw the graph of the equation x+y-1=0 and 2x+3y-12=0. Also find the vertices of a triangle formed between these two lines and x axis​

Answer 1

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

Given pair of lines are .

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

can be rewritten as

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

and

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

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

Now, we sketch the graph.

Consider equation (1),

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

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

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

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

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.

Consider equation (2)

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

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

$\rm :\longmapsto\:2(0) + 3y = 12$

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

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

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

$\rm :\longmapsto\:2x + 3(0) = 12$

$\rm :\longmapsto\:2x = 12$

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

Hᴇɴᴄᴇ,

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

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

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

➢ See the attachment graph.

From graph, we conclude that triangle ABC is the required triangle bounded by the lines with x - axis.

And the coordinates of the vertices are

Coordinates of A ( 1, 0 )

Coordinates of B ( 6, 0 )

Coordinates of C ( - 9, 10 )