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=[tex]\sqrt{(y2-y1)^2+(x2-x1)^2}[/tex]
substituting the values in the above formula
10=[tex]\sqrt{(z-(-3))^2+(-3-5)^2}[/tex]
10=[tex]\sqrt{(z+3)^2+(-8)^2}[/tex]
10=[tex]\sqrt{(z^2 +6z+9)+(64)}[/tex]
10=[tex]\sqrt{(z^2 +6z+73)}[/tex]
taking square root on both sides
100=[tex]z^2 +6z+73[/tex]
100-73=[tex]z^2 +6z[/tex]
[tex]z^2 +6z-27[/tex]=0
[tex]z^2 +9z-3z-27=0[/tex]
[tex]z(z+9) -3(z+9)=0[/tex]
[tex](z-3)(z+9)=0\\z-3=0 ; z+9=0\\z=3 : z=-9[/tex]
