Find the area of the triangle formed by the lines x + y = 6, x – 3y = 2 and 5x – 3y + 2 = 0.
The given equations are
x + y = 6 …(i)
x – 3y = 2 …(ii)
and 5x – 3y + 2 = 0
or 5x – 3y = -2 …(iii)
Let eq. (i), (ii) and (iii) represents the sides AB, BC and AC respectively of ΔABC
Firstly, we solve the equation (i) and (ii)
x + y = 6 …(i)
x – 3y = 2 …(ii)
Subtracting eq. (ii) from (i), we get
x + y – x + 3y = 6 – 2
⇒ 4y = 4
⇒ y = 1
Putting the value of y = 1 in eq. (i), we get
x + 1 = 6
⇒ x = 5
Thus, AB and BC intersect at (5, 1)
Now, we solve eq. (ii) and (iii)
x – 3y = 2 …(ii)
5x – 3y = -2 …(iii)
Subtracting eq. (ii) from (iii), we get
5x – 3y – x + 3y = – 2 – 2
⇒ 4x = - 4
⇒ x = -1
Putting the value of x = -1 in eq. (ii), we get
– 1 – 3y = 2
⇒ -3y = 2 + 1
⇒ -3y = 3
⇒ y = -1
Thus, BC and AC intersect at (-1, -1)
Now, we solve eq. (iii) and (i)
5x – 3y = -2 …(iii)
x + y = 6 …(i)
From eq. (i), we get
x = 6 – y
Putting the value of x in eq. (iii), we get
5(6 – y) – 3y = -2
⇒ 30 – 5y – 3y = -2
⇒ 30 – 8y = -2
⇒ -8y = -32
⇒ y = 4
Putting the value of y = 4 in eq. (i), we get
x + 4 = 6
⇒ x = 6 – 4
⇒ x = 2
Thus, AC and AB intersect at (2, 4)
So, vertices of triangle ABC are: (5, 1), (-1, -1) and (2, 4)
= 12 sq. units [∵, area can’t be negative]