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

The Correct Option is C

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 \((\alpha + \beta) + 2(h + k) = 47 - 42 = 5.\)