Find the slope of a line passing through P(2,3) and intersecting the line $x+y=7$ and at a distance of 4 units from P.

Question

Let the point of intersection be (h,k)

Using the symmetric form $$\frac{h-2}{\cos \theta}=4$$ $$h=4\cos \theta+2$$ and $$k=4\sin \theta+3$$ Since (h,k) lies on $x+y=7$ $$2\cos \theta+2\sin \theta-1=0$$ I found the value of $\sin \theta$ by squaring on both sides and solving the quadratic to be $$\sin \theta=\frac{1\pm\sqrt 7}{4}$$ From here I can find $\cos \theta$ and subsequently $\tan \theta=m$

What what about the $\pm$? Only one of the answer is valid, how do I find out which is the right one?

Answer 1

Another way

As the point of intersection$Q$ must lie on $x+y=7$

$Q(h,7-h)$

Now $4^2=(2-h)^2+(7-h-3)^2$

Clearly the above quadratic equation has two distinct real roots, each one corresponds to one possible position of $Q$

Answer 2

Note that the two tangents $m=\tan\theta$ corresponding to $\sin \theta=\frac{1\pm\sqrt 7}{4}$ are respectively,

$$m_1= -\frac{4+\sqrt7}3=-2.22,\>\>\>\>\>\>\>m_2 = -\frac{4-\sqrt7}3=-0.45$$

The options in your MCQ should contain one of the two solutions above. So, check $m_1$ and $m_2$ against the choices and identify the matching one.