Question

(20 points)

Answer 1

Answer:

the possible coordinates are -9 and 3

Step-by-step explanation:

coordinates of B=(x2,y2)= (5,1)

Coordinates of A= (x1, y1)= (-3, z)

distance= 10 units

Formula: d=$\sqrt{(y2-y1)^2+(x2-x1)^2}$

substituting the values in the above formula

10=$\sqrt{(z-(-3))^2+(-3-5)^2}$

10=$\sqrt{(z+3)^2+(-8)^2}$

10=$\sqrt{(z^2 +6z+9)+(64)}$

10=$\sqrt{(z^2 +6z+73)}$

taking square root on both sides

100=$z^2 +6z+73$

100-73=$z^2 +6z$

$z^2 +6z-27$=0

$z^2 +9z-3z-27=0$

$z(z+9) -3(z+9)=0$

$(z-3)(z+9)=0\\z-3=0 ; z+9=0\\z=3 : z=-9$