Understanding the Problem: Altitude of a Triangle

The problem asks for the equation of the altitude drawn from vertex B to the side AC in a triangle ABC. We are given the equations of the three sides of the triangle.

An altitude of a triangle is a line segment from a vertex that is perpendicular to the opposite side. To find the equation of the altitude through B on AC, we need two things:

1. The coordinates of vertex B.
2. The slope of the side AC (so we can find the perpendicular slope).

Finding the Coordinates of Vertex B

Vertex B is the intersection point of the sides AB and BC. We are given the equations for AB and BC:

• Equation of AB: \(x - 2 = 0\)
• Equation of BC: \(y + 1 = 0\)

From the equation of AB, we directly get \(x = 2\). From the equation of BC, we directly get \(y = -1\).

Therefore, the coordinates of vertex B are \((2, -1)\).

Finding the Slope of Side AC

The equation of side AC is given as \(x + 2y - 4 = 0\). To find the slope, we can rewrite this equation in the slope-intercept form, \(y = mx + c\), where \(m\) is the slope.

Rearrange the equation of AC:

\(x + 2y - 4 = 0\)

\(2y = -x + 4\)

Divide by 2:

\(y = -\frac{1}{2}x + 2\)

Comparing this to \(y = mx + c\), the slope of AC is \(m_{AC} = -\frac{1}{2}\).

Finding the Slope of the Altitude through B

The altitude through B on AC is perpendicular to side AC. If two lines are perpendicular, the product of their slopes is -1. Let the slope of the altitude be \(m_{altitude}\).

We have \(m_{altitude} \times m_{AC} = -1\).

\(m_{altitude} \times (-\frac{1}{2}) = -1\)

Solving for \(m_{altitude}\):

\(m_{altitude} = \frac{-1}{-\frac{1}{2}} = 2\)

So, the slope of the altitude through B on AC is 2.

Finding the Equation of the Altitude through B on AC

We know that the altitude passes through point B \((2, -1)\) and has a slope \(m_{altitude} = 2\). We can use the point-slope form of a linear equation, which is \(y - y_1 = m(x - x_1)\), where \((x_1, y_1)\) is the point and \(m\) is the slope.

Substitute the coordinates of B \((2, -1)\) and the slope \(m=2\):

\(y - (-1) = 2(x - 2)\)

\(y + 1 = 2x - 4\)

To get the equation in the standard form \(Ax + By + C = 0\), move all terms to one side:

\(2x - y - 4 - 1 = 0\)

\(2x - y - 5 = 0\)

This is the equation of the altitude through B on AC.

Verification by Comparing with Options

Let's compare our derived equation \(2x - y - 5 = 0\) with the given options:

• Option 1: \(x - 3y + 1 = 0\)
• Option 2: \(x - 3y + 4 = 0\)
• Option 3: \(2x - y + 4 = 0\)
• Option 4: \(2x - y - 5 = 0\)

Our equation matches Option 4.

The equation of the altitude through B on AC is \(2x - y - 5 = 0\).

Revision Table: Key Concepts for Altitude Equation

Additional Information: Altitudes and Orthocenter

Every triangle has three altitudes, one from each vertex perpendicular to the opposite side. The three altitudes of a triangle are concurrent, meaning they all intersect at a single point. This point of intersection is called the orthocenter of the triangle.

Finding the orthocenter involves finding the equations of at least two altitudes and solving them simultaneously. In this problem, we found the equation of one altitude. To find the orthocenter, you would need to find the equation of another altitude (e.g., from A on BC, or from C on AB) and find where the two altitudes intersect.

The concept of slope, perpendicular lines, and finding the equation of a line passing through a point with a known slope are fundamental in coordinate geometry problems involving triangles and other geometric shapes.