Question

\(\mathrm{A}(6,7), \mathrm{B}(-2,3)\) and \(\mathrm{C}(9,1)\) are vertices of \(\Delta \mathrm{ABC}\), then coordinates of point of intersection of bisector of \(\angle \mathrm{B}\) and side \(\underline{\mathrm{AC}}\) is (a) \([-(22 / 3),(13 / 3)]\) (b) \([(22 / 3),(13 / 3)]\) (c) \([(22 / 3),-(13 / 3)]\) (d) \([-(22 / 3),-(13 / 3)]\)

Short answer

The coordinates of the point of intersection of the bisector of angle B and side AC are \[\left(\frac{22}{3}, \frac{13}{3}\right)\].

Step-by-step solution

Calculate the lengths of the sides.

First, we need to calculate the lengths of the sides AB, BC, and CA of the triangle ABC using the distance formula, which is \[D = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\] AB = \(\sqrt{(-2 - 6)^2 + (3 - 7)^2}\) = 10 BC = \(\sqrt{(9 + 2)^2 + (1 - 3)^2}\) = 11 CA = \(\sqrt{(9 - 6)^2 + (1 - 7)^2}\) = 10

Apply angle bisector theorem.

The angle bisector theorem states that the ratio of the length of the line segment created by the angle bisector to the length of the remaining side is equal to the ratio of the lengths of the other two sides of the triangle. In other words: \(BD / BA = CD / CA\) Let the coordinates of D be \((x, y)\). Then, let \(BD = k\) and \(CD = l\). We have: \(k / 10 = l / 10\)

Use the section formula to find coordinates of D.

We know the ratio k : l = 10 : 11, as BD : CD = 10 : 11. We can use the section formula to find coordinates of D: \(x = \frac{kC_x + lA_x}{k + l}\) and \(y = \frac{kC_y + lA_y}{k + l}\) Solving for x and y: \(x = \frac{10 * 9 + 11 * 6}{10 + 11}\) \(y = \frac{10 * 1 + 11 * 7}{10 + 11}\)

Calculate the coordinates of D.

Now, calculate the coordinates: \(x = \frac{90 + 66}{21} = \frac{156}{21} = \frac{22}{3}\) \(y = \frac{10 + 77}{21} = \frac{87}{21} = \frac{13}{3}\) So the coordinates of D are \((\frac{22}{3}, \frac{13}{3})\). The answer is (b).

Key concepts

Coordinate Geometry

Coordinate geometry, also known as analytic geometry, combines algebra and geometry to describe the position of points, lines, and figures in a two-dimensional space using coordinates. In the context of the exercise, we're dealing with a triangle in a Cartesian plane, where each vertex is defined by an ordered pair \((x, y)\).

The beauty of coordinate geometry lies in its ability to convert geometric problems into algebraic form, making them easier to solve with equations. In this exercise, we use coordinate geometry not only to visualize the triangle ABC but also to establish a relationship between its sides and angles through algebraic expressions.

Distance Formula

The distance formula is pivotal in coordinate geometry as it allows us to calculate the distance between any two points in a Cartesian plane. The formula is derived from the Pythagorean theorem and is expressed as \(D = \sqrt{(x_2 - x_1)\^2 + (y_2 - y_1)\^2}\), where \(D\) is the distance, and \((x_1, y_1)\) and \((x_2, y_2)\) are the coordinates of the two points.

Using the Distance Formula:

In the given problem, we calculated the lengths of the sides AB, BC, and CA of triangle ABC using this formula. These distances are essential to apply the angle bisector theorem correctly as they determine the ratio on which the angle bisector divides the opposite side, Point D in this case.

Section Formula

The section formula in coordinate geometry is a tool for finding the coordinates of a point that divides a line segment into a particular ratio. There are two versions of the section formula: one for internal division and one for external division. For internal division, the formula is: \( \left(x, y\right) = \left(\frac{m x_2 + n x_1}{m + n}, \frac{m y_2 + n y_1}{m + n}\right) \), where \(m\) and \(n\) are the ratios and \( (x_1, y_1) \), \( (x_2, y_2) \) are the endpoints of the line segment.

Applying the Section Formula:

For our exercise, we use the section formula to find the coordinates of D, the intersection point of the angle bisector of \(\angle\ B \) at side \( \bar{AC} \) in triangle ABC. Knowing the proportion in which D divides AC is crucial. Using that ratio, the section formula gives us the precise location of D, thus solving the underlying geometric problem with algebra.