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]


1