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Find the locus of points R(x, y) such that the distance from R to P(2, 5) is three times the distance from R to Q(4, 7).

Mathematics

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Answer

2x2+2y2−17x−29y+139=0\mathbf{2x^2 + 2y^2 - 17x - 29y + 139 = 0}

Step-by-step solution

5 steps

  1. Define the coordinates and the given condition.

    Let the coordinates of point PP be (2,5)(2, 5), point QQ be (4,7)(4, 7), and the variable point RR be (x,y)(x, y).
    The given condition is PR=3QRPR = 3QR.

  2. Write the distance formulas for PRPR and QRQR.

    The distance formula between two points (xa,ya)(x_a, y_a) and (xb,yb)(x_b, y_b) is (xb−xa)2+(yb−ya)2\sqrt{(x_b - x_a)^2 + (y_b - y_a)^2}.

    PR=(x−2)2+(y−5)2PR = \sqrt{(x - 2)^2 + (y - 5)^2} QR=(x−4)2+(y−7)2QR = \sqrt{(x - 4)^2 + (y - 7)^2}

  3. Substitute the distance formulas into the given condition and square both sides.

    PR=3QRPR = 3QR (x−2)2+(y−5)2=3(x−4)2+(y−7)2\sqrt{(x - 2)^2 + (y - 5)^2} = 3\sqrt{(x - 4)^2 + (y - 7)^2}

    Squaring both sides:

    (x−2)2+(y−5)2=9[(x−4)2+(y−7)2](x - 2)^2 + (y - 5)^2 = 9\left[(x - 4)^2 + (y - 7)^2\right]

  4. Expand and simplify the equation.

    Expand the squared terms:

    (x2−4x+4)+(y2−10y+25)=9[(x2−8x+16)+(y2−14y+49)](x^2 - 4x + 4) + (y^2 - 10y + 25) = 9\left[(x^2 - 8x + 16) + (y^2 - 14y + 49)\right] x2+y2−4x−10y+29=9[x2+y2−8x−14y+65]x^2 + y^2 - 4x - 10y + 29 = 9\left[x^2 + y^2 - 8x - 14y + 65\right]

    Distribute the 9 on the right side:

    x2+y2−4x−10y+29=9x2+9y2−72x−126y+585x^2 + y^2 - 4x - 10y + 29 = 9x^2 + 9y^2 - 72x - 126y + 585

  5. Rearrange the terms to form the equation of the locus.

    Move all terms to one side of the equation:

    0=9x2−x2+9y2−y2−72x+4x−126y+10y+585−290 = 9x^2 - x^2 + 9y^2 - y^2 - 72x + 4x - 126y + 10y + 585 - 29 0=8x2+8y2−68x−116y+5560 = 8x^2 + 8y^2 - 68x - 116y + 556

    Divide the entire equation by 4 to simplify:

    0=8x24+8y24−68x4−116y4+55640 = \frac{8x^2}{4} + \frac{8y^2}{4} - \frac{68x}{4} - \frac{116y}{4} + \frac{556}{4} 0=2x2+2y2−17x−29y+1390 = 2x^2 + 2y^2 - 17x - 29y + 139

    The equation of the locus of RR is 2x2+2y2−17x−29y+139=0\boxed{\mathbf{2x^2 + 2y^2 - 17x - 29y + 139 = 0}}.

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