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Straight Lines and Pair of Straight Lines question

2025 · 24 Jan · Shift 1 · Q43
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  5. /2025 · 24 Jan · Shift 1 · Q43

Straight Lines and Pair of Straight Lines question

2025 · 24 Jan · Shift 1 · Q43

JEE MainMathematicsStraight Lines and Pair of Straight LinesMCQ+4 / −1
Let the lines 3x−4y−α=0,8x−11y−33=03 x-4 y-\alpha=0,8 x-11 y-33=03x−4y−α=0,8x−11y−33=0, and 2x−3y+λ=02 x-3 y+\lambda=02x−3y+λ=0 be concurrent. If the image of the point (1,2)(1,2)(1,2) in the line 2x−3y+λ=02 x-3 y+\lambda=02x−3y+λ=0 is (5713,−4013)\left(\frac{57}{13}, \frac{-40}{13}\right)(1357​,13−40​), then ∣αλ∣|\alpha \lambda|∣αλ∣ is equal to
  1. A
    91
  2. B
    113
  3. C
    101
  4. D
    84
View written solutionFree

Correct answer: A

  1. Use the reflection condition to find λ\lambdaλ.

The line is 2x−3y+λ=0.2x-3y+\lambda=0.2x−3y+λ=0. If P(1,2)P(1,2)P(1,2) is reflected in this line to P′(5713,−4013)P'\left(\frac{57}{13},-\frac{40}{13}\right)P′(1357​,−1340​), then the line is the perpendicular bisector of PP′PP'PP′.

A standard reflection formula says that for line ax+by+c=0ax+by+c=0ax+by+c=0, the image of (x1,y1)(x_1,y_1)(x1​,y1​) is

y_1-\frac{2b(ax_1+by_1+c)}{a^2+b^2}\right).$$ Here, $$a=2,\quad b=-3,\quad c=\lambda,\quad (x_1,y_1)=(1,2).$$ So $$ax_1+by_1+c=2(1)-3(2)+\lambda=\lambda-4,$$ and $$a^2+b^2=2^2+(-3)^2=13.$$ Hence the image is $$\left(1-\frac{4(\lambda-4)}{13}, 2+\frac{6(\lambda-4)}{13}\right).$$ This equals $$\left(\frac{57}{13},-\frac{40}{13}\right).$$ From the $x$-coordinate: $$1-\frac{4(\lambda-4)}{13}=\frac{57}{13}$$ $$\frac{13-4\lambda+16}{13}=\frac{57}{13}$$ $$29-4\lambda=57$$ $$-4\lambda=28$$ $$\lambda=-7.$$ Check with the $y$-coordinate: $$2+\frac{6(\lambda-4)}{13}=-\frac{40}{13}$$ $$26+6(\lambda-4)=-40$$ $$26+6\lambda-24=-40$$ $$6\lambda=-42$$ $$\lambda=-7,$$ which is consistent. --- 2. **Find the common point of the second and third lines.** The three concurrent lines are: $$3x-4y-\alpha=0,$$ $$8x-11y-33=0,$$ $$2x-3y+\lambda=0.$$ Since $\lambda=-7$, the third line is $$2x-3y-7=0.$$ Now solve $$8x-11y-33=0 \quad \Rightarrow \quad 8x-11y=33$$ and $$2x-3y-7=0 \quad \Rightarrow \quad 2x-3y=7.$$ Multiply the second equation by $4$: $$8x-12y=28.$$ Subtract from the first: $$ (8x-11y)-(8x-12y)=33-28$$ $$y=5.$$ Then $$2x-3(5)=7$$ $$2x-15=7$$ $$2x=22$$ $$x=11.$$ So the concurrency point is $$(11,5).$$ --- 3. **Use the first line to find** $\alpha$. Since $(11,5)$ lies on $$3x-4y-\alpha=0,$$ we get $$3(11)-4(5)-\alpha=0$$ $$33-20-\alpha=0$$ $$\alpha=13.$$ --- 4. **Compute** $|\alpha\lambda|$. $$|\alpha\lambda|=|13\cdot(-7)|=91.$$ --- 5. **Compare with the stored answer.** Derived answer: **91**. Stored correct answer: **A = 91**. They match.
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