Skip to content
NCERT Exemplar · Q8

Q.Find the equation of the line passing through the point of intersection of 2x+y=52x+y=5 and x+3y+8=0x+3y+8=0 and parallel to the line 3x+4y=73x+4y=7.

Punjab PsebShort· 3mImportance★★★★★
65% · 94/145 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 →

Using the family-of-lines method, the required line is 3x+4y+3=03x + 4y + 3 = 0.

Any line through the intersection of 2x+y−5=02x + y - 5 = 0 and x+3y+8=0x + 3y + 8 = 0 can be written as

(2x+y−5)+λ(x+3y+8)=0,(2x + y - 5) + \lambda(x + 3y + 8) = 0,

i.e. (2+λ)x+(1+3λ)y+(8λ−5)=0.(2+\lambda)x + (1+3\lambda)y + (8\lambda - 5) = 0.

This must be parallel to 3x+4y=73x + 4y = 7 (slope −34-\tfrac34), so the coefficients of xx and yy are in the ratio 3:43:4: …

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.