Question

If a straight line through the point P (3, 5) makes an angle $\frac{\pi}{6}$ with +ve x-axis and meets the lines 2x + y + 5 = 0 and 3x - 2y + 7 = 0 at Q and R then $\frac{PQ}{PR}$ =

Answer 1

$\frac{8(20 - 7\sqrt{3})}{33}$

Answer 2

To find the ratio $\frac{PQ}{PR}$, we will use the parametric form of a straight line.

1. Equation of the line passing through P(3, 5):

The line passes through point $P(x_1, y_1) = (3, 5)$ and makes an angle $\theta = \frac{\pi}{6}$ with the positive x-axis. The parametric equation of the line is given by:

$\frac{x - x_1}{\cos\theta} = \frac{y - y_1}{\sin\theta} = r$

Here, $\cos\theta = \cos(\frac{\pi}{6}) = \frac{\sqrt{3}}{2}$ and $\sin\theta = \sin(\frac{\pi}{6}) = \frac{1}{2}$. Substituting the values:

$\frac{x - 3}{\frac{\sqrt{3}}{2}} = \frac{y - 5}{\frac{1}{2}} = r$

From this, we can express $x$ and $y$ in terms of $r$:

$x = 3 + r\frac{\sqrt{3}}{2}$ $y = 5 + r\frac{1}{2}$

Here, $r$ represents the directed distance from point P to any point $(x, y)$ on the line. $PQ = |r_Q|$ and $PR = |r_R|$.

2. Finding the distance PQ ($r_Q$):

The line meets the first line $2x + y + 5 = 0$ at Q. Substitute the parametric expressions for $x$ and $y$ into the equation of the first line:

$2\left(3 + r_Q\frac{\sqrt{3}}{2}\right) + \left(5 + r_Q\frac{1}{2}\right) + 5 = 0$ $6 + r_Q\sqrt{3} + 5 + r_Q\frac{1}{2} + 5 = 0$ $16 + r_Q\left(\sqrt{3} + \frac{1}{2}\right) = 0$ $16 + r_Q\left(\frac{2\sqrt{3} + 1}{2}\right) = 0$ $r_Q = -\frac{16 \times 2}{2\sqrt{3} + 1} = -\frac{32}{2\sqrt{3} + 1}$

So, $PQ = |r_Q| = \frac{32}{2\sqrt{3} + 1}$.

3. Finding the distance PR ($r_R$):

The line meets the second line $3x - 2y + 7 = 0$ at R. Substitute the parametric expressions for $x$ and $y$ into the equation of the second line:

$3\left(3 + r_R\frac{\sqrt{3}}{2}\right) - 2\left(5 + r_R\frac{1}{2}\right) + 7 = 0$ $9 + r_R\frac{3\sqrt{3}}{2} - 10 - r_R + 7 = 0$ $6 + r_R\left(\frac{3\sqrt{3}}{2} - 1\right) = 0$ $6 + r_R\left(\frac{3\sqrt{3} - 2}{2}\right) = 0$ $r_R = -\frac{6 \times 2}{3\sqrt{3} - 2} = -\frac{12}{3\sqrt{3} - 2}$

So, $PR = |r_R| = \frac{12}{3\sqrt{3} - 2}$.

4. Calculating the ratio $\frac{PQ}{PR}$:

$\frac{PQ}{PR} = \frac{\frac{32}{2\sqrt{3} + 1}}{\frac{12}{3\sqrt{3} - 2}}$ $\frac{PQ}{PR} = \frac{32}{2\sqrt{3} + 1} \times \frac{3\sqrt{3} - 2}{12}$

Divide 32 by 4 and 12 by 4:

$\frac{PQ}{PR} = \frac{8}{2\sqrt{3} + 1} \times \frac{3\sqrt{3} - 2}{3}$ $\frac{PQ}{PR} = \frac{8(3\sqrt{3} - 2)}{3(2\sqrt{3} + 1)}$

To rationalize the denominator, multiply the numerator and denominator by the conjugate of $(2\sqrt{3} + 1)$, which is $(2\sqrt{3} - 1)$:

$\frac{PQ}{PR} = \frac{8(3\sqrt{3} - 2)(2\sqrt{3} - 1)}{3(2\sqrt{3} + 1)(2\sqrt{3} - 1)}$

Numerator: $8[(3\sqrt{3})(2\sqrt{3}) - (3\sqrt{3})(1) - (2)(2\sqrt{3}) + (-2)(-1)]$ $= 8[18 - 3\sqrt{3} - 4\sqrt{3} + 2]$ $= 8[20 - 7\sqrt{3}]$

Denominator: $3[(2\sqrt{3})^2 - 1^2]$ $= 3[4 \times 3 - 1]$ $= 3[12 - 1]$ $= 3[11]$ $= 33$

Therefore, $\frac{PQ}{PR} = \frac{8(20 - 7\sqrt{3})}{33}$

Explanation of the solution:

1. Use the parametric form of a straight line $\frac{x - x_1}{\cos\theta} = \frac{y - y_1}{\sin\theta} = r$, where $(x_1, y_1) = (3, 5)$ and $\theta = \frac{\pi}{6}$.
2. Substitute $x = 3 + r\frac{\sqrt{3}}{2}$ and $y = 5 + r\frac{1}{2}$ into the equation of the first line ($2x + y + 5 = 0$) to find $r_Q$.
3. Substitute $x = 3 + r\frac{\sqrt{3}}{2}$ and $y = 5 + r\frac{1}{2}$ into the equation of the second line ($3x - 2y + 7 = 0$) to find $r_R$.
4. Calculate the ratio $\frac{|r_Q|}{|r_R|}$ and simplify by rationalizing the denominator.