A line, with the slope greater than one, passes through the point A(4,3) and intersects th…

Question

A line, with the slope greater than one, passes through the point $A(4,3)$ and intersects the line $x-y-2=0$ at the point $B$. If the length of the line segment $AB$ is $\dfrac{\sqrt{29}}3$, then $B$ also lies on the line:

Accepted Answer

Let the inclination be $\theta$. $B = \left(4-\dfrac{\sqrt{29}}3\cos\theta,\ 3-\dfrac{\sqrt{29}}3\sin\theta\right)$, satisfying $x-y-2=0$:
$$4-\frac{\sqrt{29}}3\cos\theta-3+\frac{\sqrt{29}}3\sin\theta-2=0 \implies \sin\theta-\cos\theta = \frac3{\sqrt{29}}$$

Squaring: $1-\sin2\theta = \dfrac9{29} \implies \sin2\theta = \dfrac{20}{29} = \dfrac{2\tan\theta}{1+\tan^2\theta}$

Solving gives $\tan\theta = \dfrac52$ (taking the root $>1$).

So $\sin\theta=\dfrac5{\sqrt{29}},\ \cos\theta=\dfrac2{\sqrt{29}}$:
$$B = \left(4-\frac{\sqrt{29}}3\cdot\frac2{\sqrt{29}},\ 3-\frac{\sqrt{29}}3\cdot\frac5{\sqrt{29}}\right) = \left(4-\frac23,\ 3-\frac53\right) = \left(\frac{10}3,\frac43\right)$$

Check $x+2y=\dfrac{10}3+\dfrac83=6$.

Solutions · 1