Given:
x + y – 4 = 0, 3x – 7y + 8 = 0, 4x – y – 31 = 0 forming a triangle and point (-3, 2)
Let ABC be the triangle of sides AB, BC and CA, whose equations x + y − 4 = 0, 3x − 7y + 8 = 0 and 4x − y − 31 = 0, respectively.
On solving them, we get A (7, – 3), B (2, 2) and C (9, 5) as the coordinates of the vertices.
Let P (− 3, 2) be the given point.
The given point P (− 3, 2) will lie inside the triangle ABC, if
(i) A and P lies on the same side of BC
(ii) B and P lies on the same side of AC
(iii) C and P lies on the same side of AB
Thus, if A and P lie on the same side of BC, then
21 + 21 + 8 – 9 – 14 + 8 > 0
50 × – 15 > 0
-750 > 0,
This is false
∴ The point (−3, 2) lies outside triangle ABC.