91Ó°ÊÓ

The area of a triangle in coordinate geometry can be calculated through a specific formula using the vertices of the triangle. When you have a triangle with vertices

• \((x_1, y_1)\)
• \((x_2, y_2)\)
• \((x_3, y_3)\)

the area is given by \[\text{Area} = \frac{1}{2} \left| x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2) \right|. \]This formula might look a bit complex, but it's systematic. Each term represents a unique way of pairing up the vertices and taking a difference in their coordinates.
Substituting the values of the triangle's vertices into this formula allows you to neatly compute the area. In the exercise above, we substitute the points\[A(2,3), B(1,-3), C(-4,-7)\] into the formula to determine the area of triangle ABC. This results in an area of 13 square units. By applying the formula step-by-step, calculating the area becomes straightforward.

The slope of a line is an essential concept in coordinate geometry. It determines the steepness or incline of the line and is denoted by \(m\). A slope can be positive, negative, zero, or undefined. In mathematical terms, for a line passing through two points

• \((x_1, y_1)\)
• \((x_2, y_2)\)

the slope \(m\) is calculated by \[m = \frac{y_2-y_1}{x_2-x_1}. \]This formula gives you the ratio of vertical change to horizontal change between any two points on the line. In the example from our exercise, the line we consider is between points \(B(1, -3)\) and \(D(-1, -2)\).
After substituting these coordinates into the formula, the slope of line BD is determined as \(-\frac{1}{2}\), indicating that for every unit increase in \(x\), \(y\) decreases by half a unit. Understanding the slope helps in determining the line's direction and inclination, which are critical in more advanced topics such as finding perpendicular lines.

The concept of perpendicular lines is fundamental in both geometry and algebra. In coordinate geometry, a line is said to be perpendicular to another if the product of their slopes is \(-1\).
If a line has a slope of \(m\), then a line perpendicular to it will have a slope \(-\frac{1}{m}\). This relationship is particularly useful when you are asked to find the equation of a line that is perpendicular to another given line.
In the given solution, we calculated the slope of line BD as \(-\frac{1}{2}\). Thus, the slope of the line perpendicular to BD becomes \(-\frac{1}{-\frac{1}{2}}\), which equals \(2\). We utilize this perpendicular slope \(2\) alongside point \(B(1, -3)\) to find the equation for this perpendicular line. The equation, using point-slope form, becomes \(y = 2x - 5\), which can be reorganized as \(y - 2x + 5 = 0\).
This principle is vital when solving problems related to angles, perpendicular bisectors, or when constructing polygons in coordinate planes.