Question

The equation of the internal bisector of ∠BAC of ΔABC with vertices A(5, 2), B(2, 3) and C(6, 5), is

A. 2x + y + 12 = 0
B. x + 2y – 12 = 0
C. 2x + y – 12 = 0
D. 2x + y – 12 = 0

Answer 1

To find the equation of the internal bisector of angle BAC in triangle ABC, we can find the midpoint of line segment BC, calculate the slope of the internal bisector, and use the point-slope form of a line to find the equation.

Step-by-step explanation:

To find the equation of the internal bisector of angle BAC in triangle ABC, we need to find the coordinates of the point where the internal bisector intersects the line segment BC. This point is the midpoint of the line segment BC. We can find the midpoint by averaging the x-coordinates and y-coordinates of B and C.

The midpoint is ((2 + 6)/2, (3 + 5)/2) = (4, 4).

Now, we can find the slope of the internal bisector using the slope formula: (y2 - y1) / (x2 - x1). Using the coordinates of A (5, 2) and the midpoint (4, 4), we have: (4 - 2) / (4 - 5) = 2 / -1 = -2.

Since the slope of the internal bisector is -2 and it passes through the midpoint (4, 4), we can use the point-slope form of a line to find the equation of the internal bisector: y - 4 = -2(x - 4).

Simplifying this equation gives us the answer: 2x + y - 12 = 0.