Question

Two vertices of a triangle $\triangle ABC$ are $A(3, -1)$ and $B(-2, 3)$, and its orthocentre is $P(1, 1)$. If the coordinates of the point $C$ are $(\alpha, \beta)$ and the centre of the circle circumscribing the triangle $\triangle PAB$ is $(h, k)$, then the value of $(\alpha + \beta) + 2(h + k)$ equals:

• A. 51
• B. 81
• C. 5
• D. 15

Answer 1

Answer: (C)

5

Answer 2

We are given that $M_{AB} = \frac{4}{5}$ and $M_{DP} = \frac{5}{4}$.

The equation of the line $PC$ is given by:

$y - 1 = \frac{5}{4}(x - 1) \quad \text{(Equation 1)}.$

Also, we are given that $M_{AP} = \frac{-2}{-2} = -1$, which implies $M_{BC} = +1$.

The equation of line $BC$ is:

$y - 3 = (x + 2) \quad \text{(Equation 2)}.$

Now, solving Equations (1) and (2):

$x + 4 = \frac{5}{4}(x - 1) \implies 4x + 16 = 5x - 5 \implies \alpha = 21$

Next, using $\beta = y = x + 5$, we get:

$\alpha + \beta = 47$

The equation of the perpendicular bisector of $AP$ is:

$y - 0 = (x - 2) \quad \text{(Equation 3)}$

The equation of the perpendicular bisector of $AB$ is:

$y - 1 = \frac{5}{4}(x - \frac{1}{2}) \quad \text{(Equation 4)}$

Now, solving Equations (3) and (4):

$(x - 3)4 = 5x - \frac{5}{2} \implies x = -\frac{19}{2} = h$

Substitute this value of $x$ into the equation for $y$:

$y = -\frac{23}{2} = k$

Finally, we calculate:

$2(h + k) = 2 \left( -\frac{19}{2} - \frac{23}{2} \right) = -42$

Thus, the value of $(\alpha + \beta) + 2(h + k) = 47 - 42 = 5.$