Equation of BD Bisecting AC at 90° in Quadrilateral

Answer

Based on the image provided, the diagonal $BD$ acts as the perpendicular bisector of segment $AC$. By calculating the midpoint $M$ and the negative reciprocal gradient of $AC$, we find the equation of $BD$ to be $y = \frac{5}{2}x - \frac{27}{4}$.

Explanation

The diagram shows a quadrilateral $ABCD$. Point $A$ is at $(6, 1)$, $C$ is at $(1, 3)$, and $B$ lies on the $x$-axis. The diagonal $BD$ intersects $AC$ at a point $M$, bisecting it at $90^\circ$.

1. Calculate the midpoint M of AC Since $BD$ bisects $AC$, it must pass through the midpoint $M$ of $A(6,1)$ and $C(1,3)$. $M = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) = \left( \frac{6+1}{2}, \frac{1+3}{2} \right) = (3.5, 2)$ The midpoint formula provides the average coordinates to find the center of the line segment.

2. Determine the gradient of AC To find the slope of the perpendicular line $BD$, we first need the slope of $AC$. $m_{AC} = \frac{y_2 - y_1}{x_2 - x_1} = \frac{3 - 1}{1 - 6} = \frac{2}{-5} = -0.4$ The gradient formula measures the steepness or "rise over run" between two points.

3. Find the gradient of BD ⚠️ This step is required on exams. Use the perpendicular property: $m_1 \cdot m_2 = -1$. $m_{BD} = -\frac{1}{m_{AC}} = -\frac{1}{-2/5} = \frac{5}{2}$ When two lines are perpendicular, their gradients are negative reciprocals of each other.

4. Derive the equation of BD We use the point-slope form $y - y_1 = m(x - x_1)$ with point $M(3.5, 2)$ and gradient $m = 2.5$. $y - 2 = \frac{5}{2}(x - 3.5)$ Expand and simplify the equation: $y - 2 = 2.5x - 8.75$ $y = 2.5x - 6.75$ This linear equation defines all points $(x, y)$ that lie on the diagonal $BD$.

Final Answer

The equation of the line $BD$ is: $\boxed{y = \frac{5}{2}x - \frac{27}{4} \text{ or } y = 2.5x - 6.75}$

Common Mistakes

• Gradient Confusion: Students often forget to take the negative reciprocal and mistakenly use the same gradient as the bisected line.
• Wrong Point: Using point $A$ or point $C$ to find the equation of $BD$ instead of the midpoint $M$. Only $M$ is guaranteed to be on line $BD$.