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The Straight Lines — MATHS STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceMATHSThe Straight Lines3 Marks
Question
Find the equation of the straight lines passing through the origin and making an angle of 45° with the straight line $\sqrt{3}\text{x}+\text{y}=11.$
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Answer

Let the required equation be ax + by = c but here it passes through origin (0, 0) $\therefore$ c = 0 $\therefore$ Equation is ax + by = 0 Slope of the line $(\text{m}_1)=\frac{\text{-a}}{\text{b}}$ and $\text{m}_2=\frac{-\sqrt3}{1}$ ⇒ Angle between $\sqrt3\text{x+y}=11$ and $\text{ax+by}=0$ is 45° $\therefore\ \tan45^\circ=\frac{\text{m}_1\pm\text{m}_2}{1\pm\text{m}_1\text{m}_2}$ $1=\frac{\frac{\text{-a}}{\text{b}}\pm(-\sqrt3)}{1\mp\frac{\text{a}}{\text{b}}\times\sqrt3}$ $1-\frac{\sqrt3\text{a}}{\text{b}}=\frac{-\text{a}}{\text{b}}-\sqrt{3}$ and $1+\frac{\text{a}}{\text{b}}\sqrt3=\frac{-\text{a}}{\text{b}}+\sqrt3$ $\text{b}-\sqrt3\text{a}=-\text{a}-\sqrt{3}\text{b}$ and $\text{b}+\text{a}\sqrt{3}=-\text{a}+\text{b}\sqrt3$ $\text{a}(1-\sqrt{3})=\text{b}(-\sqrt3-1)$ and $\text{a}(\sqrt{3+1})=\text{b}(\sqrt{3-1})$ $\frac{\text{a}}{\text{b}}-\frac{1-\sqrt3}{\sqrt3-1}=\frac{(\sqrt3-1)^2}{2}=2-\sqrt3$ or $\frac{\text{a}}{\text{b}}=\frac{\sqrt3-1}{\sqrt3-1}=-2-\sqrt3$ $\therefore$ Required lines are $ \frac{\text{y}}{\text{x}}=\sqrt3\pm2\text{ or y}=(\sqrt3\pm2)\text{x}$

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