Question

Draw the graph of the following equation name the triangle find its coordinate and area
x=0; y=0; x-2y-3=0​

Answer 1

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

Given equation of lines are

$\rm :\longmapsto\:x - 2y - 3 = 0 - - - (1)$

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

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

Consider Equation (1),

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

can be rewritten as

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

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

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

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

$\rm :\longmapsto\: y = - 1.5$

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

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

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

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

Hᴇɴᴄᴇ,

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

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

➢ Now draw a graph using the points (0 , - 1.5) & (3 , 0)

➢ See the attachment graph.

Consider Equation 2

$\rm :\longmapsto\:x = 0 \: represents \: y \: axis$

Consider Equation 3

$\rm :\longmapsto\:y = 0 \: represents \: x\: axis$

So, from graph we concluded that

The required triangle formed by the lines x - 2y = 3, x = 0 and y = 0 is OAB having

Coordinates of O (0, 0)

Coordinates of A (3, 0)

Coordinates of B (- 1.5, 0)

Now, OA = 3 units and OB = 1.5 units

Area of triangle OAB

$\rm \: = \: \:\dfrac{1}{2} \times OA \times OB$

$\rm \: = \: \:\dfrac{1}{2} \times 3 \times 1.5$

$\rm \: = \: \:1.5 \times 1.5$

$\rm \: = \: \:2.25 \: sq. \: units$