3 years ago

Find a point on the y-axis that is equidistant from the points (7, −7) and (1, 1).

1 answer:

6 0

Hello :
a point on the y-axis is : P( 0, y)
PA = PB or PA² = PB².... A(7, -7) B(1,1)
(7-0)²+(-7-y)² = (1-0)² + (1-y)²
49+49+14y+y² = 1+1-2y+y²
16y = - 96
y= - 6

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Step-by-step explanation:

Given that; QT = 9 and TP = 8.

From the diagram, join R to P. Thus RP = QP (radius of the circle)

RP = QT + TP

= 9 + 8

RP = 17

Applying Pythagoras theorem to triangle TRP;

RT = $\sqrt{(RP)^{2} - (TP)^{2} }$

= $\sqrt{17^{2} - 8^{2} }$

= $\sqrt{225}$

= 15

∴ RT = 15

But, RT = TS = 15.

So that;

RS = 15 + 15

= 30

RS = 30

Therefore; RP = 17, RT = 15 and RS = 30.

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