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A point R moves such that 5PR = 3QR. If P=(2,5) and Q=(4,7), find the locus of R.

Mathematics

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Answer

4x2−7x+4y2−31y+35=04x^2 - 7x + 4y^2 - 31y + 35 = 0

Step-by-step solution

6 steps

  1. Define the coordinates and the given condition.

    Let the coordinates of point RR be (x,y)(x, y).
    The given fixed points are P(2,5)P(2, 5) and Q(4,7)Q(4, 7).
    The condition is 5∣PR⃗∣=3∣QR⃗∣5|\vec{PR}| = 3|\vec{QR}|.

  2. Write the distance formulas for ∣PR⃗∣|\vec{PR}| and ∣QR⃗∣|\vec{QR}|.

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

    ∣PR⃗∣=(x−2)2+(y−5)2|\vec{PR}| = \sqrt{(x-2)^2 + (y-5)^2} ∣QR⃗∣=(x−4)2+(y−7)2|\vec{QR}| = \sqrt{(x-4)^2 + (y-7)^2}

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

    5(x−2)2+(y−5)2=3(x−4)2+(y−7)25\sqrt{(x-2)^2 + (y-5)^2} = 3\sqrt{(x-4)^2 + (y-7)^2}

    Squaring both sides:

    (5(x−2)2+(y−5)2)2=(3(x−4)2+(y−7)2)2(5\sqrt{(x-2)^2 + (y-5)^2})^2 = (3\sqrt{(x-4)^2 + (y-7)^2})^2 25((x−2)2+(y−5)2)=9((x−4)2+(y−7)2)25((x-2)^2 + (y-5)^2) = 9((x-4)^2 + (y-7)^2)

  4. Expand the squared terms.

    (x−2)2=x2−4x+4(x-2)^2 = x^2 - 4x + 4 (y−5)2=y2−10y+25(y-5)^2 = y^2 - 10y + 25 (x−4)2=x2−8x+16(x-4)^2 = x^2 - 8x + 16 (y−7)2=y2−14y+49(y-7)^2 = y^2 - 14y + 49

    Substitute these expansions back into the equation:

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

  5. Distribute the constants and rearrange the terms.

    25x2−100x+25y2−250y+725=9x2−72x+9y2−126y+58525x^2 - 100x + 25y^2 - 250y + 725 = 9x^2 - 72x + 9y^2 - 126y + 585

    Move all terms to one side to form the general equation of the locus:

    (25x2−9x2)+(−100x+72x)+(25y2−9y2)+(−250y+126y)+(725−585)=0(25x^2 - 9x^2) + (-100x + 72x) + (25y^2 - 9y^2) + (-250y + 126y) + (725 - 585) = 0 16x2−28x+16y2−124y+140=016x^2 - 28x + 16y^2 - 124y + 140 = 0

  6. Simplify the equation by dividing by the common factor.

    All coefficients are divisible by 4. Divide the entire equation by 4:

    16x24−28x4+16y24−124y4+1404=0\frac{16x^2}{4} - \frac{28x}{4} + \frac{16y^2}{4} - \frac{124y}{4} + \frac{140}{4} = 0 4x2−7x+4y2−31y+35=04x^2 - 7x + 4y^2 - 31y + 35 = 0

    This is the equation of the locus of R.

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