Question

If the straight line through the point $P(3,4)$ makes an angle $\frac{\pi }{6}$ with the x-axis and meets the line $12x+5y+10=0$ at Q, then the length PQ is?

• A. $\frac{132}{12\sqrt{3}+5}$
• B. $\frac{132}{12\sqrt{3}-5}$
• C. $\frac{132}{5\sqrt{3}+12}$
• D. $\frac{132}{5\sqrt{3}-12}$

Answer 1

The correct option is A $\frac{132}{12\sqrt{3}+5}$

The tangent of $\frac{\pi }{6}=30$ degrees is $0.57735$.

The first line has a slope of $0.57735$ because tangent is the change in $y$ coordinates divided by the change in $x$ coordinates, which is also the tangent. Start with point slope form $(y-y_1=m(x-x_1\left)\right)$ and change it to slope intercept form $(y=mx+c)$

$y-4=0.57735(x-3)$

$y-4=0.57735x-1.73205$

Add 4 on both sides to isolate y.

$y=0.57735x+2.26795$

Now take the second equation isolate $y$ and then we set the two $y$ values equal to each other.

$12x+5y+10=0$

$5y=-12x-10$

Divide both sides by $5$, $y$ will now be isolated.

$y=-2.4x-2$

Now set both expressions for y equal to each other.

$ie.,0.57735x+2.26795=-2.4x-2$

$2.97735x+2.26795=-2$

$2.97735x=-4.26795$

$x=-1.43347271$…. Approximate this as $x=-1.43347$

Now find $y$ by putting in $-1.43347$ as the value of $x$

$y=-2.4(-1.43347)-2$

$\therefore y=1.440328$

So point $Q$ is $(-1.43347,1.440328)$

Now apply the distance formula $d=\sqrt{(x_2-x_1{)}^2+(y_2-y_1{)}^2}$

$=\sqrt{(3-(-1.43347){)}^2+(4-1.44347{)}^2}$

$=\sqrt{19.65565624+6.535845641}$

$=5.11776336$…. approximate as $5.11777$

Simplifying the given options we get the solution as $\frac{132}{12\sqrt{3}+5}$. Option A