Question

Find the distance of the point (3,4) from the midpoint of the line joining (8,10) and (4,6)

Answer 1

Answer:

Distance between point $(3,4)$ and midpoint of line joining $(8,10)$ and $(4,6)$ = $5$ units.

Step-by-step explanation:

Given:

Points:

$A(3,4)\\B(8,10)\\C(4,6)$

To find distance from point A to midpoint of  BC.

Midpoint M of BC:

$M=(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})$

$M=(\frac{8+4}{2},\frac{10+6}{2})$     [Plugging in points $B(8,10)\ and\ C(4,6)$]

$M=(\frac{12}{2},\frac{16}{2})$

$M=(6,8)$

Distance between A and M:

$D=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

$D=\sqrt{(6-3)^2+(8-4)^2}$   [Plugging in points $A(3,4)\ and\ M(6,8)$]

$D=\sqrt{(3)^2+(4)^2}$

$D=\sqrt{9+16}$

$D=\sqrt{25}$

$D=\pm5$

Since distance is always positive ∴ $D= 5$ units