Given equations of lines are x+2y−5=0 . . . (i) and 3x−4y+5=0 . . . (ii) and let P≡[(a−1)2,a]
From Eq. (i), origin O and P are on the same side. (−5)(a−1)2+2a−5)>0 ⇒a2+1−2a+2a−5<0 ⇒a2<4 ⇒−2<a<2 From Eq. (ii), we get (5) [3(a−1)2−4a+5]>0 ⇒3(a2+1−2a)−4a+5>0 ⇒3a2−6a+3−4a+5>0 ⇒3a2−10a+8>0 ⇒(a−2)(3a−4)>0 ⇒a<
4
3
or a>2
Now, Eqs. (iii) ∩ (iv), a∈(−2,4∕3) ∴a={−1,0,1} Hence, there are 3 points.