The values of y are 1 and 3
What is distance formula -
[tex]\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}[/tex]
x1 and y1 are x and y coordinates of a point.
similarly x2 and y2 are x and y coordinates of another point.
Explanation with steps -
As point P is equidistant from Q and R, the distances PQ and PR should be the same.
So, calculate distance PQ can be calculated from the above formula
The distance PQ is [tex]\sqrt{50}[/tex]
now use this distance to calculate the unkown 'y'
[tex]\sqrt{50} =[/tex] [tex]\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}[/tex]
[tex]\sqrt{50} =[/tex] [tex]\sqrt{(-10-(-3))^2+(y-2)^2}[/tex]
[tex]\sqrt{50} =[/tex] [tex]\sqrt{49+(y-2)^2}[/tex]
Squaring both sides
50 = 49 + [tex](y-2)^{2}[/tex]
[tex]y^{2}-4y+3=0[/tex]
On factorizing,
[tex](y-1)(y-3)=0[/tex]
Thus we get values of y to be 1 or 3
To see more about distance formula
refer : https://brainly.com/question/1872885
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