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5.1 · Q2

Q.A(−5, 2) and B(4, 1). Find the equation of the locus of point P, which is equidistant from A and B.

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✓ Free question

Let P(x,y)P(x,y) be equidistant from A(−5,2)A(-5,2) and B(4,1)B(4,1), so PA2=PB2PA^2=PB^2:

(x+5)2+(y−2)2=(x−4)2+(y−1)2.(x+5)^2+(y-2)^2=(x-4)^2+(y-1)^2.

Expanding: x2+10x+25+y2−4y+4=x2−8x+16+y2−2y+1x^2+10x+25+y^2-4y+4 = x^2-8x+16+y^2-2y+1.

Cancel x2,y2x^2,y^2: 10x−4y+29=−8x−2y+1710x-4y+29=-8x-2y+17.

Bring together: 18x−2y+12=018x-2y+12=0, divide by 2: 9x−y+6=09x-y+6=0.

✓Final answer

9x−y+6=09x - y + 6 = 0

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