Consider a triangle having vertices A(-2,3), B(1,9) and C(3,8).

Question

Consider a triangle having vertices $A(-2,3),\ B(1,9)$ and $C(3,8)$. If a line $L$ passing through the circumcentre of $	riangle ABC$, bisects line $BC$, and intersects $Y$-axis at point $\left(0,\dfrac{\alpha}2ight)$, then the value of real number $\alpha$ is:

Answer: 9

Accepted Answer

$$AB=\sqrt{9+36}=\sqrt{45}, \quad BC=\sqrt{4+1}=\sqrt5, \quad AC=\sqrt{25+25}=\sqrt{50}$$

Since $AC^2=AB^2+BC^2$ ($50=45+5$), $\angle B=90^\circ$, so the circumcentre is the midpoint of hypotenuse $AC$: $\left(\dfrac12,\dfrac{11}2\right)$.

Midpoint of $BC = \left(2,\dfrac{17}2\right)$.

Line through these two points:
$$y-\frac{11}2 = \frac{\frac{17}2-\frac{11}2}{2-\frac12}\left(x-\frac12\right) = 2\left(x-\frac12\right)$$

At $x=0$: $y=\dfrac{11}2-1=\dfrac92$.

Given this equals $\dfrac{\alpha}2$: $\alpha=9$.

Solutions · 1