JEE Mains · Maths · STD 11 - 9. straight line
A ray of light along \(x + \sqrt 3 y = \sqrt 3 \) gets reflected upon reaching \(x- \) axis , the equation of the reflected ray is
- A \(\;y = x + \sqrt 3 \)
- B \(\;\sqrt 3 y = x - \sqrt 3 \)
- C \(\;y = \sqrt 3 x - \sqrt 3 \)
- D \(\;\sqrt 3 y = x - 1\)
Answer & Solution
Correct Answer
(B) \(\;\sqrt 3 y = x - \sqrt 3 \)
Step-by-step Solution
Detailed explanation
Take any point \(\mathrm{B}(0,1)\) on given line. Equation of \(\mathrm{AB}^{\prime}\) is \( y-0=\frac{-1-0}{0-\sqrt{3}}(x-\sqrt{3}) \) \( \Rightarrow -\sqrt{3} y=-x+\sqrt{3}\) \( \Rightarrow x-\sqrt{3} y=\sqrt{3} \) \( \Rightarrow \sqrt{3} y=x-\sqrt{3} \)
See the Complete Solution
Get step-by-step explanations for this and 2.5 Lakh+ more JEE, NEET & CET questions.
- Unlock all solutions
- Practice the full chapter
- Track accuracy across PYQs
4.8 rated on Google Play · 14,000+ reviews
More questions from Maths
- If \(\mathrm{a}_{\mathrm{r}}=\cos \frac{2 \mathrm{r} \pi}{9}+i \sin \frac{2 \mathrm{r} \pi}{9}, \mathrm{r}=1,2,3, \ldots, i=\sqrt{-1}\) then the determinant \(\left|\begin{array}{lll}a_{1} & a_{2} & a_{3} \\ a_{4} & a_{5} & a_{6} \\ a_{7} & a_{8} & a_{9}\end{array}\right|\) is equal to :JEE Mains 2021 Medium
- A chord is drawn through the focus of the parabola \(y^2\, = 6x\) such that its distance from the vertex of this parabola is \(\frac{{\sqrt 5 }}{2}\) , then its slope can be:JEE Mains 2014 Hard
- Let \({f_k}\left( x \right) = \frac{1}{k}\left( {{{\sin }^k}x + {{\cos }^k}x} \right)\) where \(x \in R\;\) and \(k \ge 1\). Then \({f_4}\left( x \right) - {f_6}\left( x \right) \) is equalsJEE Mains 2014 Hard
- Let the system of linear equations \(x+y+\alpha z=2\) \(3 x+y+z=4\) \(x+2 z=1\) have a unique solution \(\left(x^{*}, y^{*}, z^{*}\right)\). If \(\left(\alpha, x^{*}\right),\left(y^{*}, \alpha\right)\) and \(\left(x^{*},-y^{*}\right)\) are collinear points, then the sum of absolute values of all possible values of \(\alpha\) isJEE Mains 2022 Hard
- Let \(\alpha_1, \alpha_2, \ldots, \alpha_7\) be the roots of the equation \(x^7+\) \(3 x^5-13 x^3-15 x=0\) and \(\left|\alpha_1\right| \geq\left|\alpha_2\right| \geq \ldots \geq\left|\alpha_7\right|\). Then \(\alpha_1 \alpha_2-\alpha_3 \alpha_4+\alpha_5 \alpha_6\) is equal to \(..................\).JEE Mains 2023 Hard
- \(\lim \limits_{x \rightarrow a} \frac{(a+2 x)^{\frac{1}{3}}-(3 x)^{\frac{1}{3}}}{(3 a+x)^{\frac{1}{3}}-(4 x)^{\frac{1}{3}}}(a \neq 0)\) is equal toJEE Mains 2020 Hard
More PYQs from JEE Mains
- Consider the lines \(\mathrm{L}_1: \mathrm{x}-1=\mathrm{y}-2=\mathrm{z}\) and \(\mathrm{L}_2: \mathrm{x}-2=\mathrm{y}=\mathrm{z}-1\). Let the feet of the perpendiculars from the point \(\mathrm{P}(5,1,-3)\) on the lines \(\mathrm{L}_1\) and \(\mathrm{L}_2\) be \(Q\) and \(R\) respectively. If the area of the triangle PQR is A , then \(4 \mathrm{~A}^2\) is equal to :JEE Mains 2025 Hard
- \( (\frac{1}{3}+\frac{4}{7})+(\frac{1}{3^{2}}+\frac{1}{3}\times\frac{4}{7}+\frac{4^{2}}{7^{2}})+(\frac{1}{3^{3}}+\frac{1}{3^{2}}\times\frac{4}{7}+\frac{1}{3}\times\frac{4^{2}}{7^{2}}+\frac{4^{3}}{7^{3}}) + \dots \) upto infinite terms is equal to -JEE Mains 2026 Easy
- Let \(A=\sum_{i=1}^{10} \sum_{j=1}^{10} \min \{i, j\}\) and \(B=\sum_{i=1}^{10} \sum_{j=1}^{10}\max \{i, j\}\). Then \(A+B\) is equal toJEE Mains 2022 Hard
- If \(A = \left[ {\begin{array}{*{20}{c}}
{\cos \,\theta }&{ - \sin \,\theta }\\
{\sin \,\theta }&{\cos \,\theta }
\end{array}} \right]\), then the matrix \({A^{ - 50}}\) when \(\theta = \frac{\pi }{{12}}\) is equal toJEE Mains 2019 Hard - Statement \(1\) : The only circle having radius \(\sqrt {10} \) and a diameter along line \(2x + y = 5\) is \(x^2 + y^2 - 6x +2y = 0\).
Statement \(2\) : \(2x + y = 5\) is a normal to the circle \(x^2 + y^2 -6x+2y = 0\).JEE Mains 2013 Hard - If the two lines \(l_{1}: \frac{ x -2}{3}=\frac{ y +1}{-2}, z =2\) and \(l_{2}: \frac{x-1}{1}=\frac{2 y+3}{\alpha}=\frac{z+5}{2}\) perpendicular, then an angle between the lines \(l_{2}\) and \(l_{3}: \frac{1- x }{3}=\frac{2 y -1}{-4}=\frac{ z }{4}\) isJEE Mains 2022 Medium