Question

PQ is a straight line of 13 units if p has the coordinates (5,-3) and q has the coordinates (-7,y) find the values of y​

Answer 1

Answer:

y = -2

Step-by-step explanation:

let,p(5,-3) = (X1, y1)

Q(-7,y) = (x2,y2)

by distance formula

d(P,Q) =√(x2 - X1)^2 + (y2 - y1)^2

13 = √(-7-5)^2 + (y-(-3)^2

13 = √(-12)^2 + (y+3)^2

13 = √144 + (y+3)^2

13 = 12 + y + 3

1 = y + 3

y = -2