What is the distance between (−3, 1
) and (1, 4
) on the coordinate plane?
Responses
5√
units
square root of 5 units
13−−√
units
square root of 13 units
5
units
5 units
13
units
To find the distance between two points on a coordinate plane, we can use the distance formula.
The distance formula is:
d = √((x2 - x1)^2 + (y2 - y1)^2)
Given the points (-3, 1) and (1, 4), we can substitute the values into the formula:
d = √((1 - (-3))^2 + (4 - 1)^2)
Simplifying, we get:
d = √((4)^2 + (3)^2)
= √(16 + 9)
= √(25)
= 5
Therefore, the distance between the two points is 5 units.
To find the distance between two points in the coordinate plane, you can use the distance formula. The distance formula is derived from the Pythagorean theorem and calculates the length of the straight line connecting the two points.
The distance formula is:
d = √((x2 - x1)^2 + (y2 - y1)^2)
In this case, the coordinates of the first point are (-3, 1) and the coordinates of the second point are (1, 4).
Using the distance formula, we can calculate the distance as follows:
d = √((1 - (-3))^2 + (4 - 1)^2)
= √((1 + 3)^2 + (3)^2)
= √((4)^2 + (3)^2)
= √(16 + 9)
= √(25)
= 5 units
Therefore, the distance between (-3, 1) and (1, 4) on the coordinate plane is 5 units.
To find the distance between two points on a coordinate plane, you can use the distance formula. The distance formula is given by:
d = √((x2 - x1)^2 + (y2 - y1)^2)
where (x1, y1) and (x2, y2) are the coordinates of the two points.
In this case, the coordinates are (-3, 1) and (1, 4).
Plugging in the values into the distance formula, we get:
d = √((1 - (-3))^2 + (4 - 1)^2)
= √((1 + 3)^2 + (4 - 1)^2)
= √((4)^2 + (3)^2)
= √(16 + 9)
= √(25)
= 5
So, the distance between (-3, 1) and (1, 4) on the coordinate plane is 5 units.