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Exercise 6.3 · Q6

Q.Find the equation of the lines passing through the point of intersection of the lines 4x−y+3=04x - y + 3 = 0 and 5x+2y+7=05x + 2y + 7 = 0, and

(i) through the point (−1,2)(-1, 2)
(ii) parallel to x−y+5=0x - y + 5 = 0
(iii) perpendicular to x−2y+1=0x - 2y + 1 = 0.
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Use the family L1+λL2=0L_1+\lambda L_2=0 with L1=4x−y+3L_1{=}4x-y+3, L2=5x+2y+7L_2{=}5x+2y+7; solve λ\lambda per condition.

  • (i) through (−1,2)(-1,2): λ=12⇒x+1=0\lambda=\tfrac12 \Rightarrow x+1=0.

  • (ii) parallel to x−y+5=0x-y+5=0: λ=−37⇒x−y=0\lambda=-\tfrac37 \Rightarrow x-y=0.

  • (iii) perpendicular to x−2y+1=0x-2y+1=0: λ=−6⇒2x+y+3=0\lambda=-6 \Rightarrow 2x+y+3=0.

Every line through the intersection of L1≡4x−y+3=0L_1\equiv 4x-y+3=0 and L2≡5x+2y+7=0L_2\equiv 5x+2y+7=0 can be written as L1+λL2=0L_1+\lambda L_2=0 for some real λ\lambda — this avoids computing the intersection point directly. Each part just imposes one further condition to pin down λ\lambda.

Step 1. Set up the family.

(4x−y+3)+λ(5x+2y+7)=0  ⟹  (4+5λ)x+(−1+2λ)y+(3+7λ)=0.(∗)(4x-y+3)+\lambda(5x+2y+7)=0 \;\Longrightarrow\; (4+5\lambda)x+(-1+2\lambda)y+(3+7\lambda)=0. \qquad (\ast)

Step 2 (i). Through the point (−1,2)(-1,2). Substitute x=−1, y=2x=-1,\ y=2 directly into the unexpanded family:

(4(−1)−2+3)+λ(5(−1)+2(2)+7)=0  ⟹  (−3)+λ(6)=0  ⟹  λ=12.\big(4(-1)-2+3\big)+\lambda\big(5(-1)+2(2)+7\big)=0 \implies (-3)+\lambda(6)=0 \implies \lambda=\tfrac12.

Substitute into (∗)(\ast) and clear the fraction (multiply by 22):

8x−2y+6+5x+2y+7=0  ⟹  13x+13=0  ⟹  x+1=0.8x-2y+6+5x+2y+7=0 \implies 13x+13=0 \implies x+1=0.

Step 3 (ii). Parallel to x−y+5=0x-y+5=0. From (∗)(\ast), the coefficients are a=4+5λ, b=−1+2λa=4+5\lambda,\ b=-1+2\lambda. Parallel to x−y+5=0x-y+5=0 (i.e. a2=1,b2=−1a_2=1,b_2=-1) requires a⋅(−1)=b⋅1a\cdot(-1)=b\cdot1, i.e. −(4+5λ)=(−1+2λ)-(4+5\lambda)=(-1+2\lambda): …

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