Question

triangle ABC has vertices at A(1,-3),B(3,0), and C(5,-3). what is the distance from vertex B to the midpoint of AC

Answer 1

Answer:

The distance from vertex B to the midpoint of AC is 3.

Step-by-step explanation:

From Linear Algebra we understand that location of the midpoint of AC is determined by the following formula:

$M(x,y) = \frac{1}{2}\cdot A(x,y) + \frac{1}{2}\cdot C(x,y)$ (1)

Where:

$A(x,y)$, $C(x,y)$ - Locations of vertices A and C regarding origin, dimensionless.

$M(x,y)$ - Location of the midpoint regarding origin, dimensionless.

If we know that $A(x,y) = (1,-3)$ and $C(x,y) = (5,-3)$, then the midpoint of AC is:

$M(x,y) = \frac{1}{2}\cdot (1,-3)+\frac{1}{2}\cdot (5,-3)$

$M(x,y) = \left(\frac{1}{2},-\frac{3}{2} \right)+\left(\frac{5}{2}, -\frac{3}{2} \right)$

$M(x,y) = (3, -3)$

Lastly, the distance from vertex B to the midpoint of AC is calculated from the Pythagorean Theorem:

$d = \sqrt{(M_{x}-B_{x})^{2}+(M_{y}-B_{y})^{2}}$ (2)

If we know that $M_{x} = 3$, $M_{y} = -3$, $B_{x} = 3$ and $B_{y} = 0$, then the distance is:

$d = \sqrt{(3-3)^{2}+(-3-0)^{2}}$

$d = 3$

The distance from vertex B to the midpoint of AC is 3.