Question

Math geometry question 4, Thanks if you help

Answer 1

The distance between two points J and K is 13 unit (approx)

Step-by-step explanation:

Given,

Two points are J(3,0) and k(6,-13).

To find the distance between J and K.

Formula

The distance between two points ($x_{1} ,y_{1}$) and ($x_{2} ,y_{2}$) is $\sqrt{(x_{2}-x_{1} ) ^{2}+(y_{2}-y_{1} ) ^{2} }$

Now,

Putting, $x_{1} = 3, y_{1}=0, x_{2}=6, y_{2}=-13$ we get,

JK = $\sqrt{(6-3)^{2} +(-13-0)^{2} }$

= $\sqrt{178}$  unit = 13.34 unit (approx)

Answer 2

JDK-right triangle. JD=13, DK=3

By the Pythagorean theorem

$13^2+3^2=JK^2\\JK=\sqrt{169+9} \\JK=\sqrt{178}$

JK≈13,34

Answer: the distance is 13 units

P.S. Hello from Russia