The vertices of triangle ABC are A(-5, 7), B(-1, -5) and C(1, 3) as shown in the diagram - Scottish Highers Maths - Question 1 - 2015

To find the median from point B to the midpoint of AC, we first need to calculate the midpoint of AC.

1. Calculate the Midpoint of AC:

   • Coordinates of A are (-5, 7) and C are (1, 3).
   • The midpoint (M) is given by:
     $M = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) = \left( \frac{-5 + 1}{2}, \frac{7 + 3}{2} \right) = \left( -2, 5 \right).$

2. Calculate the Gradient of Median from B to M:

   • Coordinates of B are (-1, -5) and M are (-2, 5).
   • The gradient of the median ($m_{BM}$) is: $m_{BM} = \frac{y_2 - y_1}{x_2 - x_1} = \frac{5 + 5}{-2 + 1} = \frac{10}{-1} = -10.$

3. Equation of the Median:

   • Using the point-slope form through point B: $y - (-5) = -10(x - (-1))$
   • This simplifies to: $y + 5 = -10(x + 1)$ $y + 5 = -10x - 10$ $y = -10x - 15.$

To find the intersection of the altitude from C and the median from B, we can set the equations equal to each other:

1. Equations to Solve:

   • The equation from the altitude is: $x - 3y = 4,$
   • The equation from the median is: $y = -10x - 15.$

2. Substitute the Median Equation into the Altitude Equation:

   • Substitute $y$ from the median into the altitude: $x - 3(-10x - 15) = 4$
   • Simplifying this gives: $x + 30x + 45 = 4$ $31x + 45 = 4$ $31x = 4 - 45$ $31x = -41$ $x = -\frac{41}{31}.$

3. Find y-coordinate:

   • Substitute $x$ back into the median equation: $y = -10(-\frac{41}{31}) - 15$ $y = \frac{410}{31} - \frac{465}{31} = -\frac{55}{31}.$

4. Coordinates of Intersection:

   • The coordinates of the intersection point are: $\left( -\frac{41}{31}, -\frac{55}{31} \right).$