if the distance from p to the points (3,4) and (-3,4) are in the ratio 3:2 find the locus of p​

Question

if the distance from p to the points (3,4) and (-3,4) are in the ratio 3:2 find the locus of p​

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Evelyn 2 weeks 2021-10-04T15:22:12+00:00 1 Answer 0 views 0

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    2021-10-04T15:23:52+00:00

    Answer:

    The locus of P is

    5x^2+5y^2-34x+120y+29=05×2+5y2−34x+120y+29=0

    Step-by-step explanation:

    Step-by-step explanation:

    Formula used:

    The distance between two points (x_1,y_1)\:and\:(x_2,y_2)\:is\:d= \sqrt{(x_1-x_2)^2+(y_1-y_2)^2}(x1,y1)and(x2,y2)isd=(x1−x2)2+(y1−y2)2

    Let the moving point be P(h,k)

    Let the given points be A(5,-4) and B(7,6)

    Given:

    AP:BP = 2:3

    \frac{AP}{BP}=\frac{2}{3}BPAP=32

    3\:AP=2\:BP3AP=2BP

    3\:\sqrt{(h-5)^2+(k+4)^2}=2\:\sqrt{(h-7)^2+(k-6)^2}3(h−5)2+(k+4)2=2(h−7)2+(k−6)2

    squaring on both sides we get

    9\:[(h-5)^2+(k+4)^2]=4\:[(h-7)^2+(k-6)^2]9[(h−5)2+(k+4)2]=4[(h−7)2+(k−6)2]

    9\:[h^2+25-10h+k^2+16+8k]=4\:[h^2+49-14h+k^2+36-12k]9[h2+25−10h+k2+16+8k]=4[h2+49−14h+k2+36−12k]

    9h^2+225-90h+9k^2+144+72k=4h^2+196-56h+4k^2+144-48k9h2+225−90h+9k2+144+72k=4h2+196−56h+4k2+144−48k

    9h^2+225-90h+9k^2+72k=4h^2+196-56h+4k^2-48k9h2+225−90h+9k2+72k=4h2+196−56h+4k2−48k

    5h^2+5k^2-34h+120k+29=05h2+5k2−34h+120k+29=0

    \therefore∴ The locus of P is

    5x^2+5y^2-34x+120y+29=05×2+5y2−34x+120y+29=0

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