Find the distance between points p (-1,5) and q (3,4)

Distance = 4.123

Step-by-step 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

98 points djdjsi zisksis sjsjsi sisksj sksjdj sksks sjsisj

i believe it is c.y = 9x

rise over run. 9 units up over 1 unit right or 9/1