Find the numerical value of the distance between (k+2,w) and (k-1, w+4).?

Answer 1

distance = sqrt[(x2 -x1)^2 + (y2 - y1)^2 ]
d = sqrt[(k-1-k-2)^2 + (w + 4 - w)^2]
d = sqrt[(-3)^2 + 4^2]
d = sqrt[9 + 16]
d = sqrt25
d = 5

Answer 2

We use the distance formula, which is:

Distance = sqrt{(x2 - x1)^2 + (y2 - y1)^2}

Let D = distance.

Let k - 1 = x2

Let k+2 = x1

Let w + 4 = y2

Let w = y1

We now plug into the formula above and simplify.

D = sqrt{(k - 1 - (k + 2))^2 + (w + 4 - w)}

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

D = sqrt{9 + 16}

D = sqrt{25}

D = 5

Answer 3

The line segment joining the points forms the hypotenuse of a right triangle.
length of vertical leg = |(w+4)-(w)| = 4
length of horizontal leg = |(k-1)-(k+2)| = 3
hypotenuse² = 4² + 3² = 25
hypotenuse = √25 = 5

Distance between (k+2,w) and (k-1, w+4) is 5
http://i233.photobucket.com/albums/ee195/DWRead/quagmire77.jpg

Answer 4

sqrt( (k+2-k+1)^2 + (w-w-4)^2)
=sqrt( 9 + 16 )
=sqrt(25)
= 5