Respuesta :
Answer:
The possible co-ordinates of the point Q are (0,5) and (0,17).
Step-by-step explanation:
Given:
P is a point (8,11)
Q is point on y-axis
PQ = 10 units
To find co-ordinates of point Q.
Solution:
Any point on y-axis is given as [tex](0,y)[/tex] as [tex]x=0[/tex] at y-axis.
Let the point Q be = [tex](0,y)[/tex]
We use the distance formula to find length of PQ.
By distance formula:
The distance between two points [tex](x_1,y_1)[/tex] and [tex](x_2,y_2)[/tex] is given as:
[tex]d=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}[/tex]
Thus, for the point P(8,11) and Q(0,[tex]y[/tex]) the distance PQ can be given as:
[tex]PQ=\sqrt{(8-0)^2+(11-y)^2}[/tex]
[tex]PQ=\sqrt{(8)^2+(11-y)^2}[/tex]
[tex]PQ=\sqrt{64+(11-y)^2}[/tex]
Substituting PQ=10 units.
[tex]10=\sqrt{64+(11-y)^2}[/tex]
Squaring both sides.
[tex]10^2=(\sqrt{64+(11-y)^2})^2[/tex]
[tex]100=64+(11-y)^2[/tex]
Subtracting both sides by 64.
[tex]100-64=64-64+(11-y)^2[/tex]
[tex]36=(11-y)^2[/tex]
Taking square root both sides.
[tex]\sqrt{36}=\sqrt{(11-y)^2}[/tex]
[tex]\pm6=11-y[/tex]
So, we have two equations to solve:
[tex]6=11-y[/tex] and [tex]-6=11-y[/tex]
Adding [tex]y[/tex] both sides.
[tex]6+y=11-y+y[/tex] and [tex]-6+y=11-y+y[/tex]
[tex]6+y=11[/tex] and [tex]-6+y=11[/tex]
Subtracting both sides by 6 for one equation and adding 6 both sides for the other equation.
[tex]6-6+y=11-6[/tex] and [tex]-6+6+y=11+6[/tex]
∴ [tex]y=5[/tex] and [tex]y=17[/tex]
Thus, the possible co-ordinates of the point Q are (0,5) and (0,17).
