Find the values of α so that the point P(α2,α) lies inside or on the triangle formed by the lines x−5y+6=0, x−3y+2=0 and x−2y−3=0
Let ABC be the triangle of the equations whose sides AB, BC and CA are respectively.
x−5y+6=0, x−3y+2=0 and x−2y−3=0
On solving equations, we get
A(9, 3), B(4, 2) and C(13, 5)
If the point P(α,α2) lies n side the ΔABC, then
(i) A and P must be on the same side of BC
(ii) B and P must be on the same side of AC
(iii) C and P must be on the same side of AB.
Now,
A and P are on the same side of BC if
{9(1)+3(−3)+2}(α2−3α+2)>0
(9−9+2)(α2−3α+2)>0
α2−3α+2>0
(α−1)(α−2)>0
αϵ(−∞,1)v(2,∞)……(i)
B and P will lie on the same side of CA, if
{13(1)+5(−5)+6}(α2−5α+6)>0
⇒(−6)(α2−5α+6)<0
⇒α2−5α+6<0
⇒(α−2)(α−3)<0
⇒αϵ(2,3)……(ii)
C and P will lie on the same side of AB, if
{4(1)+2(−2)−3}(α2−2α−3)>0
(−3)(α2−2α−3)>0
α2−2α−3<0
(α−3)(α+1)<0
αϵ(−1,3)……(iii)
From (i), (ii), (iii)
αϵ[2,3]