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The vertices of a triangle are A(-1,-7),...

The vertices of a triangle are `A(-1,-7)`, `B(5,1)` and `C(1,4).` The equation of the internal bisector of the angle `angle ABC` is

A

`3x-7y-8=0`

B

`x-7y+2=0`

C

`3x-3y-7=0`

D

None of these

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The correct Answer is:
To find the equation of the internal bisector of the angle \( \angle ABC \) formed by the points \( A(-1,-7) \), \( B(5,1) \), and \( C(1,4) \), we can follow these steps: ### Step 1: Find the slopes of lines AB and BC The slope of line \( AB \) can be calculated using the formula: \[ m_{AB} = \frac{y_2 - y_1}{x_2 - x_1} = \frac{1 - (-7)}{5 - (-1)} = \frac{8}{6} = \frac{4}{3} \] The slope of line \( BC \) is: \[ m_{BC} = \frac{y_2 - y_1}{x_2 - x_1} = \frac{4 - 1}{1 - 5} = \frac{3}{-4} = -\frac{3}{4} \] ### Step 2: Check if the lines are perpendicular To check if lines \( AB \) and \( BC \) are perpendicular, we can multiply their slopes: \[ m_{AB} \cdot m_{BC} = \frac{4}{3} \cdot -\frac{3}{4} = -1 \] Since the product of the slopes is \(-1\), the lines are indeed perpendicular. ### Step 3: Find the angle bisector Since the lines are perpendicular, the angle bisector will have a slope that is the average of the slopes of the two lines. We can use the formula for the angle bisector slope \( m \): \[ \tan \theta = \frac{m_1 + m_2}{1 - m_1 m_2} \] Here, \( m_1 = \frac{4}{3} \) and \( m_2 = -\frac{3}{4} \). Calculating: \[ \tan \theta = \frac{\frac{4}{3} - \frac{3}{4}}{1 + \left(\frac{4}{3} \cdot -\frac{3}{4}\right)} \] Finding a common denominator for the numerator: \[ = \frac{\frac{16 - 9}{12}}{1 - 1} = \frac{\frac{7}{12}}{0} \] This indicates that we have an undefined slope, which means the angle bisector is vertical. ### Step 4: Find the equation of the angle bisector Since the angle bisector is vertical, it will have the form \( x = k \). To find \( k \), we can use the coordinates of point \( B(5,1) \) as it lies on the angle bisector. Therefore, the equation of the angle bisector is: \[ x = 5 \] ### Step 5: Write the final equation The equation of the internal bisector of angle \( ABC \) is: \[ x - 5 = 0 \]
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