Question

Draw the graph of the straight line given by the equation 4x - 3y + 36 = 0
Calculate the area of the triangle formed by the line drawn and the co-ordinate axes.

Answer 1

4x - 3y + 36 = 0
⇒ 4x - 3y = -36
⇒ -3y = -36 - 4x
⇒ 3y = 36 + 4x
⇒ y = `(36 + 4x)/(3)`

When x = - 6,
y = `(36 + 4 xx (-6))/(3)`
= `(36 - 24)/(3)`
= 4

When x = - 3,
y = `(36 + 4 xx (-3))/(3)`
= `(36 - 12)/(3)`
= 8

When x = - 9,
y = `(36 + 4 xx (-9))/(3)`
= `(36 - 36)/(3)`
= 0

X	- 9	- 3	- 6
Y	0	8	4

Plotting these points we get the required graph as shown below:

The straight line cuts the co-ordinates axis at A(0,12) and B(-9,0).
∴ The triangle ΔAOB is formed.
Area of the triangle AOB
= `(1)/(2)` x AO x OB

= `(1)/(2)` x 12 x 9
 = 54 sq . units
∴ Area of the triangle is 54 sq . units