Question

Find the area of the triangle, the equations of whose sides are y = x, y = 2x and y – 3x = 4.

Answer 1

The given equations are

y = x …(i)

y = 2x …(ii)

and y – 3x = 4 …(iii)

Let eq. (i), (ii) and (iii) represents the sides AB, BC and AC respectively of ΔABC

From eq. (i) and (ii), we get x = 0 and y = 0

Thus, AB and BC intersect at (0, 0)

Solving eq. (ii) and (iii), we get

y = 2x …(ii)

and y – 3x = 4 …(iii)

Putting the value of y = 2x in eq. (iii), we get

2x – 3x = 4

⇒ - x = 4

⇒ x = - 4

Putting the value of x = -4 in eq. (ii), we get

y = 2(- 4)

⇒ y = - 8

Thus, BC and AC intersect at (- 4, - 8)

Now, Solving eq. (iii) and (i), we get

y – 3x = 4 …(iii)

and y = x …(i)

Putting the value of y = x in eq. (iii), we get

x – 3x = 4

⇒ - 2x = 4

⇒ x = - 2

Putting the value of x = -2 in eq. (i), we get

y = - 2

Thus, AC and AB intersect at (- 2, - 2)

So, vertices of triangle ABC are: (0, 0), (- 4, - 8) and (- 2, - 2)

= 4 sq. units