COMEDK Exam 2015 Solved Paper

Given, equation of lines are
‌‌

x

a

+‌

y

b

‌=1 . . . (i)
‌‌ and ‌‌‌

x

b

+‌

y

a

‌=1 . . . (ii)
⇒bx+ay‌=ab . . . (iii)
‌‌ and ‌‌ax+by‌=ab . . . (iv)
Solving Eqs. (iii) and (iv), we get
(a2−b2)y=a2b−ab2=ab(a−b)⇒y=‌

ab

a+b

Substituting the value of y in Eq. (iii), we get
bx+a(‌

ab

a+b

)=ab⇒bx=ab−‌

a2b

a+b

⇒bx=‌

ab2

a+b

⇒x=‌

ab

a+b

∴ Point of intersection is (‌

ab

a+b

,‌

ab

a+b

).
Since, equation of the line passing through origin is y=mx
∴ When it passes through (‌

ab

a+b

,‌

ab

a+b

), then we get m=1
Hence, required equation of line is y−x=0.