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The Straight Lines — MATHS STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceMATHSThe Straight Lines4 Marks
Question
Find the perpendicular distance from the origin of the perpendicular from the point (1, 2) upon the straight line $\text{x}-\sqrt{3}\text{y}+4=0.$
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Answer

The perpendicular of (1, 2) on the straight line $\text{x}-\sqrt{3}\text{y}-4$ Then, the equation is $\text{y}-\text{y}_1=\text{m}'(\text{x}-\text{x}_1)$ $\text{x}_1=1, \ \text{y}_1=2, \ \text{m}=\frac{1}{\sqrt{3}}, \ \text{m}'=-\sqrt{3}$ $\text{y}-2=-\sqrt{3}(\text{x}-1)$ $\text{y}+\sqrt{3}\text{x}-(2+\sqrt{3})=0 \ ...(\text{i})$ The perpendicular distance from (0, 0) to (i) is $\frac{|\text{ax}_1+\text{by}_1+\text{c}|}{\sqrt{\text{a}^2+\text{b}^2}}$ $\text{a}=\sqrt{3}, \ \text{b}=1, \ \text{c}=-(2+\sqrt{3})$ $\text{x}_1=0, \ \text{y}_1=0$ $=\frac{\big|\sqrt{3}(0)+1(0)+(-2-\sqrt{3})\big|}{\sqrt{(3)^2+(1)^2}}=\frac{2+\sqrt{3}}{2}$

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