The coordinates of the end points of a line segment PQ are P(3,7) and Q (11,-6).Find the coordinates of the point R on the y axis such that PR = QR
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P(3,7) Q(11,-6)
We have the point R on the y-axis,
R (0,y)
Such that:
PR = QR
PR= sqrt[(3-0)^2 + (7-y)^2]
QR= sqrt[(11-0)^2 + (-6-y)^2
sqrt[(9+(7-y)^2]= sqrt[121 + (-6-y)^2]
Square both sides:
9+ (7-y)^2 = 121 + (-6-y)^2
Open brackets:
9 + 49 -14y + y^2 = 121+36 +12y + y^2
58 -14y = 157 + 12y
26y = -99
y= -99/26
R (0, -99/26)
kholarehman:
this isnt the answer
plz check if u hv written correct values in the question
yes these are the correct values
ok
now correct???
This is correct answer (-99/26) i have also find the same answer
thnx
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