Pis a point whose ordinate and abscissa are
same. Q = (7, 11). If length of PQ = 20, find the
co-ordinates of P.
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Answer:
given Q (7,11) PQ = d=20 units x = y
to find P (x,y)
formula used d = √(x2-x1)^2 + (y2-y1)^2
20 = √ (x-7)^2 + (x-11)^2
20 = √ x^2 + 49 - 14x + x^2 + 121 - 22x
20 = √2x^2 - 36x + 170
squaring both sides,
400 = 2x^2 - 36x + 170
2x^2 - 36x - 230 = 0
x^2 - 18x - 115 = 0
x^2 - 23x + 5x - 115 = 0
x(x - 23) + 5 (x - 23) =0
(x + 5) (x - 23) = 0
x + 5 = 0 and x - 23 = 0
x = -5 and x = 23
coordinates of P (-5,-5) or P(23,23)
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