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Help me with this thx

Help me with this thx-example-1

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Answer:

Explanation:

In order to find the coordinates of P and Q, we have to find the intersection of the circle and the line. Do this by subbing in y=x-4 into the circle equation for y:


(x-3)^2+((x-4)+2)^2=25 becomes


(x-3)^2+(x-2)^2=25 and then FOIL all that out to get


x^2-6x+9+x^2-4x+4=25. Combine like terms to get


2x^2-10x-12=0 and factor to get the zeros of

x = 6 and x = -1.

When x = 6, y = 2; when x = -1, y = -5. That answers part a.

The length of PQ is found then using the distance formula:


d=√((6-(-1))^2+(2-(-5))^2) to get


d=√(98) which, in decimal form, i 9.89949937. That answers part b.

The perpendicular bisector requires that we find the midpoint of PQ:


M=((x_2+x_1)/(2),(y_2+y_1)/(2)) which, for us, is


M=((6-1)/(2),(2-5)/(2))=(2.5, -1.5)

The perpendicular slope to the given line is the opposite reciprocal of the one given, so the perpendicular slope is -1. The equation for the perpendicular bisector of PQ goes through the midpoint with the slope of -1:


y+1.5=-1(x-2.5) and

y = -x + 1 is the perpendicular bisector of PQ.

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