The Straight Lines — MATHS STD 11 Science — Question
Gujarat BoardEnglish MediumSTD 11 ScienceMATHSThe Straight Lines4 Marks
Question
Find the equation of a line perpendicular to the line $3x - y + 5 = 0$ and at a distance of $3$ units from the origin.
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Answer
Any line perpendicular to line $\sqrt{3}\text{x}-\text{y}+5=0$ will have the slope $\frac{-1}{\text{m}}$ Where, $\text{m}\Rightarrow\text{y}=\text{mx}+\text{c}$
$\text{y}=\sqrt{3}\text{x}+5$
$\text{m}=\sqrt{3}$ Point is $(x_1y_1) = (3, 3) \text{y}-\text{y}_1=\frac{-1}{\text{m}}(\text{x}-\text{x}_1)$
$\text{y}-3=\frac{-1}{\sqrt{3}}(\text{x}-3)$
$\text{x}+\sqrt{3}\text{y}+6=0$ Point can be (-3, -3) Then, equation is $\text{x}+\sqrt{3}\text{y}-6$
$\therefore \ \text{x}+\sqrt{3}\text{y}\pm6$
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