Question
If $P$ is a point on the circle $x^2 + y^2 = 4$, $Q$ is a point on the straight line $5x + y + 2 = 0$ and $x - y + 1 = 0$ is the perpendicular bisector of $PQ$, then $13$ times the sum of abscissa of all such points $P$ is __________.
Problem Setup
Let the coordinates of point $P$ be $(x_P, y_P)$ and point $Q$ be $(x_Q, y_Q)$.
- $P$ lies on the circle $x^2 + y^2 = 4$. So, $x_P^2 + y_P^2 = 4$.
- $Q$ lies on the line $5x + y + 2 = 0$. So, $5x_Q + y_Q + 2 = 0$.
- The line $L: x - y + 1 = 0$ is the perpendicular bisector of the segment $PQ$.
Perpendicular Bisector Properties
The perpendicular bisector property implies two conditions:
- Midpoint Condition: The midpoint $M$ of $PQ$, $M = \left( \frac{x_P + x_Q}{2}, \frac{y_P + y_Q}{2} \right)$, lies on the line $L$. $ \left( \frac{x_P + x_Q}{2} \right) - \left( \frac{y_P + y_Q}{2} \right) + 1 = 0 $ $ x_P + x_Q - y_P - y_Q + 2 = 0 $ $ x_Q - y_Q = -x_P + y_P - 2 \quad \quad (1) $
- Perpendicularity Condition: The segment $PQ$ is perpendicular to the line $L$. The slope of $L$ is $m_L = 1$. The slope of $PQ$ is $m_{PQ} = \frac{y_Q - y_P}{x_Q - x_P}$. $ m_{PQ} \times m_L = -1 $ $ \frac{y_Q - y_P}{x_Q - x_P} \times 1 = -1 $ $ y_Q - y_P = -(x_Q - x_P) $ $ x_Q + y_Q = x_P + y_P \quad \quad (2) $
Expressing Q in Terms of P
Solve the system of linear equations (1) and (2) for $x_Q$ and $y_Q$:
- Adding (1) and (2): $2x_Q = (-x_P + y_P - 2) + (x_P + y_P) \implies 2x_Q = 2y_P - 2 \implies x_Q = y_P - 1$.
- Subtracting (1) from (2): $2y_Q = (x_P + y_P) - (-x_P + y_P - 2) \implies 2y_Q = 2x_P + 2 \implies y_Q = x_P + 1$.
Thus, the coordinates of $Q$ are $(y_P - 1, x_P + 1)$.
Finding Possible Points P
Substitute the coordinates of $Q$ into the equation of the line $5x + y + 2 = 0$:
$ 5(y_P - 1) + (x_P + 1) + 2 = 0 $ $ 5y_P - 5 + x_P + 1 + 2 = 0 $ $ x_P + 5y_P - 2 = 0 $This gives a linear relationship between $x_P$ and $y_P$. Now, $P(x_P, y_P)$ must satisfy both this equation and the circle equation $x_P^2 + y_P^2 = 4$.
From the linear equation, $x_P = 2 - 5y_P$. Substitute this into the circle equation:
$ (2 - 5y_P)^2 + y_P^2 = 4 $ $ 4 - 20y_P + 25y_P^2 + y_P^2 = 4 $ $ 26y_P^2 - 20y_P = 0 $ $ 2y_P(13y_P - 10) = 0 $The possible values for $y_P$ are $y_P = 0$ and $y_P = \frac{10}{13}$.
Find the corresponding $x_P$ values using $x_P = 2 - 5y_P$:
- If $y_P = 0$, then $x_P = 2 - 5(0) = 2$. Point $P_1 = (2, 0)$.
- If $y_P = \frac{10}{13}$, then $x_P = 2 - 5\left(\frac{10}{13}\right) = 2 - \frac{50}{13} = \frac{26 - 50}{13} = -\frac{24}{13}$. Point $P_2 = \left(-\frac{24}{13}, \frac{10}{13}\right)$.
Both points satisfy $x_P^2 + y_P^2 = 4$.
Calculating the Final Value
The abscissas (x-coordinates) of the possible points $P$ are $x_{P1} = 2$ and $x_{P2} = -\frac{24}{13}$.
Sum of the abscissas:
$ \text{Sum} = x_{P1} + x_{P2} = 2 + \left(-\frac{24}{13}\right) = \frac{26 - 24}{13} = \frac{2}{13} $The question asks for $13$ times the sum of the abscissas:
$ 13 \times \text{Sum} = 13 \times \left(\frac{2}{13}\right) = 2 $Was this answer helpful?
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