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

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

Explanation:

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

`M(x,y) = (1)/(2)\cdot A(x,y) + (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) = (1)/(2)\cdot (1,-3)+(1)/(2)\cdot (5,-3)`

`M(x,y) = \left((1)/(2),-(3)/(2) \right)+\left((5)/(2), -(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.