The distance of the point (2, 3) from the line 2x – 3y + 28 = 0, measured parallel to the line , is equal to [2024]

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Solution

Q.
The distance of the point (2, 3) from the line 2x – 3y + 28 = 0, measured parallel to the line $\sqrt{3} x - y + 1 = 0$ , is equal to [2024]

1 $4 + 6 \sqrt{3}$

2 $4 \sqrt{2}$

3 $6 \sqrt{3}$

4 $3 + 4 \sqrt{2}$

Ans.
(1)

Slope of a line passing through (2, 3) and parallel to line $\sqrt{3} x - y + 1 = 0$ is same as that of line $\sqrt{3} x - y + 1 = 0$

So, slope of line = $\sqrt{3}$

$\Rightarrow \tan θ = \sqrt{3} \Rightarrow \sin θ = \frac{\sqrt{3}}{2}, \cos θ = \frac{1}{2}$

So, equation of line passing through (2, 3) and having slope $\sqrt{3}$ in normal form is,

$\frac{x - 2}{1 / 2} = \frac{y - 3}{\sqrt{3} / 2} = r \Rightarrow x = \frac{r}{2} + 2, y = \frac{\sqrt{3}}{2} r + 3$

This line passes through 2x – 3y + 28 = 0

$∴ 2 (\frac{r}{2} + 2) - 3 (\frac{\sqrt{3}}{2} r + 3) + 28 = 0 \Rightarrow (\frac{2 - 3 \sqrt{3}}{2}) r = - 23$

$\Rightarrow r = \frac{- 23 \times 2}{2 - 3 \sqrt{3}} \times \frac{2 + 3 \sqrt{3}}{2 + 3 \sqrt{3}} \Rightarrow r = 2 (2 + 3 \sqrt{3}) = 4 + 6 \sqrt{3}$

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