Skip to content
Miscellaneous Exercise 4 (Solve) · Q74

Q.Find the joint equation of lines: passing through the point (3,2)(3,2), one of which is parallel to the line x−2y=2x - 2y = 2 and other is perpendicular to the line y=3y = 3.

Maharashtra MsbshseTextbookSubjectiveImportance★★★★★
53% · 74/139 Questions
🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

Parallel to x−2y=2x-2y=2 (slope 1/21/2) through (3,2)(3,2): x−2y+1=0x-2y+1=0. Perpendicular to y=3y=3 (horizontal) is vertical through (3,2)(3,2): x−3=0x-3=0. Combined: (x−2y+1)(x−3)=0(x-2y+1)(x-3)=0, expanding to …

Unlock everything free for 14 days

  • Full step-by-step solutions
  • Concept-first explanations
  • Methods, shortcuts & mistakes
  • PYQ mapping + timed mock tests

Full access for 14 days. No credit card required.