Skip to content
Exercise 6.4 · Q12

Q.Find the value of kk, if the following equation represents a pair of straight lines. Further, find whether these lines are parallel or intersecting: 12x2+7xy−12y2−x+7y+k=012x^2 + 7xy - 12y^2 - x + 7y + k = 0.

Tamil Nadu DgeTextbookSubjectiveImportance★★★★★
48% · 62/129 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 →

Apply the factorisability condition abc+2fgh−af2−bg2−ch2=0abc+2fgh-af^2-bg^2-ch^2=0, solve for kk; then test h2−abh^2-ab to decide parallel vs intersecting.

Step 1. Read off coefficients from 12x2+7xy−12y2−x+7y+k=012x^2+7xy-12y^2-x+7y+k=0.

a=12,h=72,b=−12,g=−12,f=72,c=k.a=12,\quad h=\tfrac72,\quad b=-12,\quad g=-\tfrac12,\quad f=\tfrac72,\quad c=k.

Step 2. Apply the factorisability condition.

abc=12(−12)k=−144k.abc=12(-12)k=-144k.

2fgh=2(72)(−12)(72)=−494.2fgh=2\left(\frac72\right)\left(-\frac12\right)\left(\frac72\right)=-\frac{49}{4}.

af2=12(72)2=147,bg2=−12(12)2=−3,ch2=k(72)2=49k4.af^2=12\left(\frac72\right)^2=147,\qquad bg^2=-12\left(\frac12\right)^2=-3,\qquad ch^2=k\left(\frac72\right)^2=\frac{49k}{4}.

Step 3. Combine.

−144k−494−147−(−3)−49k4=0.-144k-\frac{49}{4}-147-(-3)-\frac{49k}{4}=0.

Group kk-terms: −144k−49k4=−625k4-144k-\tfrac{49k}{4}=-\tfrac{625k}{4}. Group constants: −494−147+3=−6254-\tfrac{49}{4}-147+3=-\tfrac{625}{4}. So

−625k4−6254=0 ⟹ −6254(k+1)=0 ⟹ k=−1.-\frac{625k}{4}-\frac{625}{4}=0\ \Longrightarrow\ -\frac{625}{4}(k+1)=0\ \Longrightarrow\ k=-1.

Step 4. Decide parallel vs intersecting using h2−abh^2-ab.

h2−ab=(72)2−(12)(−12)=494+144=6254≠0.h^2-ab=\left(\frac72\right)^2-(12)(-12)=\frac{49}{4}+144=\frac{625}{4}\neq0. …

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.