Question

The coordinates of the vertices of triangle ABC are 1 -1 B 1, 4 and C 8, 4. WHat is the length in units of the line segment that connects vertex A and vertex B

Answer 1

Answer:

$AB = 5$

Step-by-step explanation:

Given

$A =(1,-1)$

$B = (1,4)$

$C = (8,4)$

Required

Distance between vertices A and B

To do this, we simply calculate distance AB using distance formula.

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

So, we have:

$AB = \sqrt{(1-1)^2 + (4 - -1)^2}$

Simplify the expression the bracket

$AB = \sqrt{0^2 + 5^2}$

Evaluate all squares

$AB = \sqrt{0 + 25}$

$AB = \sqrt{25}$

Take positive square root

$AB = 5$

Hence, the line segment that connects A and B is 5 units long.