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Q.A straight line through Q(3,2)Q(\sqrt{3}, 2) makes an angle of π6\dfrac{\pi}{6} with the positive direction of the X-axis. If the straight line intersects the line 3x−4y+8=0\sqrt{3}x - 4y + 8 = 0 at PP, then find the distance of PQPQ.

Telangana TsbieTelangana Board of Intermediate Education (Intermediate 1st Year) 2019Subjective· 4mImportance★★★★★
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Write points on the line through QQ using the parametric (distance) form, substitute into the given line's equation, and solve for the parameter r=PQr=PQ.

A line through Q(3,2)Q(\sqrt3, 2) at angle θ=π6\theta=\dfrac{\pi}{6} with the positive xx-axis has parametric form (points at signed distance rr from QQ):

x=3+rcos⁡π6=3+r32,y=2+rsin⁡π6=2+r2x = \sqrt3 + r\cos\frac{\pi}{6} = \sqrt3 + \frac{r\sqrt3}{2}, \qquad y = 2 + r\sin\frac{\pi}{6} = 2+\frac{r}{2}

This point PP lies on 3x−4y+8=0\sqrt3 x - 4y+8=0. Substituting:

3(3+r32)−4(2+r2)+8=0\sqrt3\left(\sqrt3+\frac{r\sqrt3}{2}\right) - 4\left(2+\frac{r}{2}\right)+8 = 0 …

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