The equations of perpendicular bisectors of the sides AB and AC of a `DeltaABC` are `x - y + 5 = 0` and `x+2y = 0`, respectively. If the point A is (1, - 2) the equation of the line BC is

Options

A. `23x + 14y - 40`
B. `23x + 14y + 40 =0`
C. `14x + 23y - 40 = 0`
D. `14x + 23y + 40=0`

Answer

To find the equation of the line BC in triangle ABC, we will follow these steps:

### Step 1: Identify the given information
We have the equations of the perpendicular bisectors of sides AB and AC:
1. Perpendicular bisector of AB: \(x - y + 5 = 0\)
2. Perpendicular bisector of AC: \(x + 2y = 0\)

We also know the coordinates of point A: \(A(1, -2)\).

### Step 2: Find the intersection point of the perpendicular bisectors
To find the coordinates of point M (the midpoint of BC), we need to solve the equations of the two perpendicular bisectors simultaneously.

1. From the first equation, rearranging gives:
\[ y = x + 5 \]

2. Substitute \(y\) in the second equation:
\[ x + 2(x + 5) = 0 \]
\[ x + 2x + 10 = 0 \]
\[ 3x + 10 = 0 \implies x = -\frac{10}{3} \]

3. Substitute \(x\) back into the equation for \(y\):
\[ y = -\frac{10}{3} + 5 = -\frac{10}{3} + \frac{15}{3} = \frac{5}{3} \]

Thus, the intersection point M (midpoint of BC) is:
\[ M\left(-\frac{10}{3}, \frac{5}{3}\right) \]

### Step 3: Find the slope of line AB
The slope of line AB can be determined from the perpendicular bisector of AB. The slope of the line \(x - y + 5 = 0\) is 1 (since it can be rewritten as \(y = x + 5\)). Therefore, the slope of line AB is the negative reciprocal:
\[ \text{slope of AB} = -1 \]

### Step 4: Use point-slope form to find the equation of line AB
Using point A(1, -2) and the slope of AB (-1):
\[ y - (-2) = -1(x - 1) \]
\[ y + 2 = -x + 1 \]
\[ x + y + 1 = 0 \quad \text{(Equation of line AB)} \]

### Step 5: Find the slope of line AC
The slope of line AC can be determined from the perpendicular bisector of AC. The slope of the line \(x + 2y = 0\) is -\(\frac{1}{2}\) (rewritten as \(y = -\frac{1}{2}x\)). Therefore, the slope of line AC is the negative reciprocal:
\[ \text{slope of AC} = 2 \]

### Step 6: Use point-slope form to find the equation of line AC
Using point A(1, -2) and the slope of AC (2):
\[ y - (-2) = 2(x - 1) \]
\[ y + 2 = 2x - 2 \]
\[ 2x - y - 4 = 0 \quad \text{(Equation of line AC)} \]

### Step 7: Find the equation of line BC
Now we have the coordinates of point M and the slopes of lines AB and AC. The slope of line BC can be determined using the coordinates of points B and C, which are not known yet, but we can use the midpoint M to find the equation.

Using point M \(\left(-\frac{10}{3}, \frac{5}{3}\right)\) and the slope of line BC, which is the average of the slopes of lines AB and AC:
\[ \text{slope of BC} = \frac{-1 + 2}{2} = \frac{1}{2} \]

Using point-slope form for line BC:
\[ y - \frac{5}{3} = \frac{1}{2}\left(x + \frac{10}{3}\right) \]
Multiplying through by 6 to eliminate fractions:
\[ 6y - 10 = 3x + 10 \]
Rearranging gives:
\[ 3x - 6y + 20 = 0 \]

### Final Equation of Line BC
The equation of line BC is:
\[ 3x - 6y + 20 = 0 \]

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