1.

The graphs of the equations 3x - 20y - 2 = 0 and 11x - 5y + 61 = 0 intersect at P(a, b). What is the value of (a2 + b2 - ab)/(a2 - b2 + ab)?1. \(\frac{41}{31}\)2. \(\frac{31}{41}\)3. \(\frac{37}{35}\)4. \(\frac{5}{7}\)

Answer» Correct Answer - Option 2 : \(\frac{31}{41}\)

GIVEN:

3x - 20y - 2 = 0 and 11x - 5y + 61 = 0

CALCULATION:

3x - 20y - 2 = 0 .......(1)

⇒ (11x - 5y + 61 = 0) × 4  ......(2)

On subtracting  these two equation

⇒ 41x = - 246

⇒ x = - 6

putting the value of x = - 6 in any of the two equation, we will get 

⇒ y = -1

both the equations intersect each other at point P(a,b) = P(x. y)

⇒ a = -6 and b = -1

⇒ (a2 + b2 - ab)/(a2 - b2 + ab) = (36 + 1 - 6)/(36 - 1 + 6)

⇒ 31/41

∴  (a2 + b2 - ab)/(a2 - b2 + ab)  = 31/41


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