Question

4. Find the locus of the point whose distance from the point(1, -2) is double the distance from the point (-3,5).​

Answer 1

Answer:

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Step-by-step explanation:

$so \: let \: the \\ locus \: of \: points \: (1, - 2) \: \: and \: \: ( - 3,5) \: be \: \\ x \: and \: y \: \\ \\ we \: know \: that \\ distance \: formula = \\ \sqrt{(x2 - x1) {}^{2} + (y2 - y1) {}^{2} } \\ \\ therefore \\ \: according \: to \: first \: condition \\ \\ \sqrt{(x - 1) {}^{2} + (y - ( - 2)) {}^{2} } = 2.( \sqrt{(x - ( - 3)) {}^{2} + (y - 5) {}^{2} } \\ \\ squaring \: both \: sides \\ we \: have \: then \\ \\ (x - 1) {}^{2} + (y + 2) {}^{2} = 4((x + 3) {}^{2} +( y - 5) {}^{2} \\ \\ x {}^{2} + 1 - 2x + y {}^{2} + 4 + 4y = 4(x {}^{2} + 9 + 6x + y { }^{2} + 25 - 10y) \\ \\ x {}^{2} + y {}^{2} - 2x + 4y + 5 = 4( {x}^{2} + y {}^{2} + 6x - 10y + 34) \\ \\ x {}^{2} + y {}^{2} - 2x + 4y + 5 = 4 {x}^{2} + 4 {y}^{2} + 24x - 40y + 136 \\ \\ (4x {}^{2} - x {}^{2} ) + (4y {}^{2} - y {}^{2} ) + (24x + 2x) - 40y - 4y + 136 - 5 \\ \\ 3x {}^{2} + 3y {}^{2} + 26x - 44y + 131 = 0$