X = -1-3 = -4.
Y = -4 -7 = -11.
d^2 = X^2+Y^2 = (-4)^2 + (-11)^2 = 137.
d =
Point R(3,7) and point S(−1,−4) are located within the coordinate plane.
What is the distance between the two points?
137
194−−−√
13−−√
137−−−√
could someone help me understand please
3 answers
the points are 4-units apart in the x-direction
... and 11-units apart in the y-direction
the axes are at right angles to each other
... so you can use Pythagoras to find the distance
(total distance)^2 = (x-distance)^2 + (y-distance)^2
d^2 = 4^2 + 11^2 = 16 + 121
... and 11-units apart in the y-direction
the axes are at right angles to each other
... so you can use Pythagoras to find the distance
(total distance)^2 = (x-distance)^2 + (y-distance)^2
d^2 = 4^2 + 11^2 = 16 + 121
its 137−−−√ I just finished the exam.
5 answers