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Straight Lines and Pair of Straight Lines question

2022 · 30 Jun · Shift 1 · Q31
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Straight Lines and Pair of Straight Lines question

2022 · 30 Jun · Shift 1 · Q31

JEE MainMathematicsStraight Lines and Pair of Straight LinesMCQ+4 / −1
Let α\alphaα 1, α\alphaα 2 (α\alphaα 1 2) be the values of α\alphaα fo the points (α\alphaα, −-− 3), (2, 0) and (1, α\alphaα) to be collinear. Then the equation of the line, passing through (α\alphaα 1, α\alphaα 2) and making an angle of π3{\pi \over 3}3π​ with the positive direction of the x-axis, is :
  1. A
    x−3y−33+1=0x - \sqrt 3 y - 3\sqrt 3 + 1 = 0x−3​y−33​+1=0
  2. B
    3x−y+3+3=0\sqrt 3 x - y + \sqrt 3 + 3 = 03​x−y+3​+3=0
  3. C
    x−3y+33+1=0x - \sqrt 3 y + 3\sqrt 3 + 1 = 0x−3​y+33​+1=0
  4. D
    3x−y+3−3=0\sqrt 3 x - y + \sqrt 3 - 3 = 03​x−y+3​−3=0
View written solutionFree

Correct answer: THE STORED ANSWER B APPEARS INCONSISTENT WITH THE QUESTION AS WRITTEN., IF THE STATEMENT IS EXACTLY 'PASSING THROUGH $(\ALPHA_1,\ALPHA_2)$', THEN NONE OF THE OPTIONS IS CORRECT AND THE LINE IS $$\SQRT{3}X-Y-3\SQRT{3}-1=0$$., IF THERE IS A TYPO AND THE INTENDED POINT IS $(\ALPHA_2,\ALPHA_1)$, THEN OPTION **B** IS CORRECT.

  1. Use the collinearity condition

The three points are: A(α,−3),B(2,0),C(1,α)A(\alpha,-3),\quad B(2,0),\quad C(1,\alpha)A(α,−3),B(2,0),C(1,α)

For collinearity, the area of the triangle formed by these points must be zero:

\alpha & -3 & 1\\ 2 & 0 & 1\\ 1 & \alpha & 1 \end{vmatrix}=0$$ Expanding, $$\alpha(0-\alpha)-(-3)(2-1)+1(2\alpha-0) = 0$$ $$-\alpha^2+3+2\alpha=0$$ $$\alpha^2-2\alpha-3=0$$ Factorizing: $$\alpha^2-2\alpha-3=(\alpha-3)(\alpha+1)=0$$ So, $$\alpha_1=3,\quad \alpha_2=-1$$ 2. **Find the point through which the required line passes** The line passes through $(\alpha_1,\alpha_2)$, i.e. $$(3,-1)$$ 3. **Use the given angle to find slope** A line making angle $\frac{\pi}{3}$ with the positive x-axis has slope $$m=\tan\frac{\pi}{3}=\sqrt{3}$$ 4. **Equation of the line through $(3,-1)$ with slope $\sqrt{3}$** Using point-slope form: $$y-(-1)=\sqrt{3}(x-3)$$ $$y+1=\sqrt{3}x-3\sqrt{3}$$ $$\sqrt{3}x-y-3\sqrt{3}-1=0$$ This can also be written as $$x-\frac{1}{\sqrt{3}}y-3-\frac{1}{\sqrt{3}}=0$$ which does **not** match the options directly, so let us compare carefully with the given choices. 5. **Check options** Option B is: $$\sqrt{3}x-y+\sqrt{3}+3=0$$ This line has slope $\sqrt{3}$, but substituting $(3,-1)$: $$\sqrt{3}(3)-(-1)+\sqrt{3}+3=4\sqrt{3}+4\neq 0$$ So B does **not** pass through $(3,-1)$. Now check if perhaps the intended point was $(\alpha_2,\alpha_1)=(-1,3)$. Then the required line would be: $$y-3=\sqrt{3}(x+1)$$ $$\sqrt{3}x-y+\sqrt{3}+3=0$$ which is exactly **Option B**. 6. **Conclusion** From the statement as written, $(\alpha_1,\alpha_2)=(3,-1)$, giving the line $$\sqrt{3}x-y-3\sqrt{3}-1=0$$ which is not among the options. Hence the printed/stored answer B matches only if the question intended the point to be $(\alpha_2,\alpha_1)$ instead of $(\alpha_1,\alpha_2)$.
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