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
its 137−−−√ I just finished the exam.
X = -1-3 = -4.
Y = -4 -7 = -11.
d^2 = X^2+Y^2 = (-4)^2 + (-11)^2 = 137.
d =
To find the distance between two points in the coordinate plane, you can use the distance formula. The formula is:
d = √[(x2 - x1)² + (y2 - y1)²]
In this case, let's label Point R as (x1, y1) and Point S as (x2, y2).
Point R: (x1, y1) = (3, 7)
Point S: (x2, y2) = (-1, -4)
Now substitute the coordinates into the formula:
d = √[(-1 - 3)² + (-4 - 7)²]
= √[(-4)² + (-11)²]
= √[16 + 121]
= √137
Therefore, the distance between Point R and Point S is √137.
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