Equation of line L1 passing through (2,1) and (3,‌
5
2
) y−1=‌
‌
5
2
−1
3−2
(x−2) ⇒‌‌L1⇒3x−2y=4 . . . (i) ∵‌L2⟂L1 ⇒‌‌L2⇒2x+3y ⇒‌‌L2⇒2x+3y+λ=0 It passes through (4,−1). ∴‌‌2(4)+3(−1)+λ=0 ⇒‌‌λ=−5 L2⇒2x+3y−5=0 . . . (ii) Solving Eqs. (i) and (ii), we get x=‌
22
13
‌ and ‌y=‌
7
13
Point of intersection of L1 and L2 at x=0, from Eq. (i), we get y=−2 From Eq. (ii), we get y=‌
5
3
Area of triangle formed by vertices (0,−2),(0,‌