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1 vote
Find the distance between points P (-1,5) and Q (3,4)

2 Answers

4 votes

Answer:

Explanation:

the formular for distance between two points is:
√((x2-x1)^2+(y2-y1)^2) \\\\\sqrt{4^(2)+1^(2) } \\√(17)\\ = 4.12

User Rasheena
by
5.1k points
1 vote

Answer: Distance = 4.123

Explanation:

The given coordinates are P(-1,5) and Q (3,4)

- 1 = x1 = the horizontal coordinate(along the x axis) at P

3 = x2 = the horizontal coordinate (along the x axis) at Q

5 = y1 = vertical coordinate( along the y axis) at P

4 = y2 = vertical coordinate(along the y axis) at Q

The distance between points P and Q is expressed as square root of the sum of the square of the horizontal distance and the square of the vertical distance. It becomes

Distance = √(x2 - x1)^2 + (y2 - y1)^2

Distance = √(3 - - 1)^2 + (4 - 5)^2

Distance = √ 4^2 + (-1)^2

Distance = √16+1 = √17

Distance = 4.123

User Lukasz Czerwinski
by
5.4k points