Verified answer

Answer:

[tex]y=-2[/tex]

Explanation:

Given that P(3, -2) and Q(7, y)

Distance between P and Q is 4 units

PQ = 4 units

Using distance formula [tex]\sqrt {(x_2-x_1)^2+(y_2-y_1)^}[/tex]

[tex]\sqrt{(7-3)^2+(y-(-2))^2}=4[/tex]

[tex]\sqrt{4^2+(y+2)^2}=4[/tex]

[tex]\sqrt{16+y^2+4+4y}=4\ \ \ [\because (a+b)^2=a^2+b^2+2ab][/tex]

[tex]\sqrt{y^2+4y+20}=4[/tex]

Squaring both sides

[tex]y^2+4y+20=16[/tex]

[tex]y^2+4y+4=0[/tex]

Splitting the middle term

[tex]y^2+2y+2y+4=0[/tex]

[tex]y(y+2)+2(y+2)=0[/tex]

[tex](y+2)(y+2)=0[/tex]

[tex]y+2=0,\ y+2=0[/tex]

So [tex]y=-2[/tex]