MHT CET · Maths · Pair of Lines
The joint equation of pair of lines through the origin and making an equilateral triangle with the line \(y=5\) is
- A \(3 x^2-y^2=0\)
- B \(5 x^2-y^2=0\)
- C \(x^2-3 y^2=0\)
- D \(\sqrt{3} x^2-y^2=0\)
Answer & Solution
Correct Answer
(A) \(3 x^2-y^2=0\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & y=\tan 60^{\circ} x \Rightarrow y=x \\ & y=\tan 120^{\circ} x \Rightarrow y=-x \\ & \text { joint equation }=(x-y)(x+y)=0 \\ & \Rightarrow 3 x^2-y^2=0\end{aligned}\)
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
- The value of \(\tan ^{-1}\left(\frac{1}{8}\right)+\tan ^{-1}\left(\frac{1}{2}\right)+\tan ^{-1}\left(\frac{1}{5}\right)\) isMHT CET 2023 Easy
- If the function f defined on \(\left(\frac{\pi}{6}, \frac{\pi}{3}\right)\) by
\(f(x)=\left\{\begin{array}{cc}
\frac{\sqrt{2} \cos x-1}{\cot x-1}, & x \neq \frac{\pi}{4} \
\mathrm{k}, & x=\frac{\pi}{4}
\end{array}\right.\)
is continuous, then k is equal toMHT CET 2024 Medium - If \(\mathrm{P}(\mathrm{A})=\frac{2}{5}, \mathrm{P}(\mathrm{B})=\frac{1}{4}\) and \(\mathrm{P}(\mathrm{AUB})=\frac{1}{2}\), then \(\mathrm{P}\left(\mathrm{A}^{\prime} \cup \mathrm{B}^{\prime}\right)=\)MHT CET 2020 Easy
- \(\int_{0}^{1} \frac{x^{2}}{1+x^{2}} d x=\)MHT CET 2020 Medium
- The shaded part of the given figure indicates the feasible region. Then the constraints are
MHT CET 2021 Easy - \(\int_0^{\frac{\pi}{4}} \log \left(\frac{\sin x+\cos x}{\cos x}\right) d x=\)MHT CET 2024 Medium
More PYQs from MHT CET
- Which nitrogen atom of purine base is bonded with furanose forming nucleoside?MHT CET 2022 Medium
- _________ is the most convenient and cheap method of artificial vegetative propagation.MHT CET 2016 Medium
- Formation of oogonia in human females is completed in __________ .MHT CET 2021 Hard
- Let \(\bar{a}=\alpha \hat{i}+3 \hat{\mathrm{j}}-\hat{\mathrm{k}}, \overline{\mathrm{b}}=3 \hat{i}-\hat{\mathrm{j}}+\beta \hat{\mathrm{k}}\) and \(\overline{\mathrm{c}}=\hat{i}+2 \hat{\mathrm{j}}-2 \hat{\mathrm{k}}\) where \(\alpha, \beta \in \mathbb{R}\), be three vectors. If the projection of \(\bar{a}\) on \(\bar{c}\) is \(\frac{10}{3}\) and \(\overline{\mathrm{b}} \times \overline{\mathrm{c}}=-6 \hat{i}+10 \hat{\mathrm{j}}+7 \hat{\mathrm{k}}\), then the value of \((\alpha+\beta)\) is equal toMHT CET 2025 Medium
- How many oxidation steps occur during one turn of Krebs Cycle?MHT CET 2024 Hard
- Two coherent sources ' \(\mathrm{P}\) ' and ' \(\mathrm{Q}\) ' produce interference at point ' \(\mathrm{A}\) ' on the screen, where there is a dark band which is formed between \(4^{\text {th }}\) and \(5^{\text {th }}\) bright band. Wavelength of light used is \(6000 Å\). The path difference PA and QA isMHT CET 2021 Medium