Question

Let A(-4,3), B(0,5) and C(-1,2) be three points. The equation of the straight line which passes through C and bisects the straight-line segment AB is

• A. \(y + 2x = 0\)
• B. \(y - 2x + 2x = 0\)
• C. \(y + 2x - 4 = 0\)
• D. \(y + 2x + 4 = 0\)
• E. \(y - 2x = 0\)

Hint:

Whenever a line passes through the origin, its constant term \(c\) in \(Ax + By + c = 0\) is zero. Notice that point \((0,0)\) satisfies \(y + 2x = 0\), which is a quick way to verify this type of equation.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
The required line passes through point C and the midpoint of segment AB. We first calculate the midpoint and then use the two-point form to find the line's equation.

Step 2: Key Formula or Approach:
1. Midpoint formula: \(M = \left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)\).
2. Slope: \(m = \frac{y_2-y_1}{x_2-x_1}\).
3. Line equation: \(y - y_1 = m(x - x_1)\).

Step 3: Detailed Explanation:

Step 1: Find the midpoint \(M\) of AB.
\[ M = \left(\frac{-4 + 0}{2}, \frac{3 + 5}{2}\right) = (-2, 4) \]

Step 2: The line passes through \(C(-1, 2)\) and \(M(-2, 4)\).
Find the slope \(m\):
\[ m = \frac{4 - 2}{-2 - (-1)} = \frac{2}{-1} = -2 \]

Step 3: Find the equation using point \(C(-1, 2)\):
\[ y - 2 = -2(x - (-1)) \]
\[ y - 2 = -2(x + 1) \]
\[ y - 2 = -2x - 2 \]
\[ y + 2x = 0 \]

Step 4: Final Answer:
The equation of the line is \(y + 2x = 0\).