Question

the line 2x-3y-12=0 intersects X and Y axes at A and B respectively. find the area of triangle AOB

Answer 1

Given:

A line $2x-3y-12=0$ intersecting the X and Y axes at points A and B respectively.

To find:

Area of $\triangle AOB$.

Solution:

First of all, let us find the coordinates of A and B.

A is the point on X axis, therefore $y$ will be 0.

Putting the value of $y$ as 0 in the given equation.

$2x - 3(0) - 12 = 0\\\Rightarrow x = 6$

Point A will be (6, 0).

B is the point on Y axis, therefore $x$ will be 0.

Putting the value of $y$ as 0 in the given equation.

$2(0) -3y -12 =0\\\Rightarrow y = -4$

Point B will be (0, -4).

Kindly refer to the attached image in the answer area.

Clearly, Distance OB = 4 units and distance OA = 6 units

Also, the given lines are perpendicular and base of the right angled $\triangle AOB$.

Formula for area of right angled triangle:

$Area = \dfrac{1}{2}\times Base\times Height\\ Area of \triangle AOB = \dfrac{1}{2}\times 4\times 6 = \bold{12\ sq\ units}$