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QuestionFind the equation of the locus of point P(x,y)P(x, y) such that the distances to points S(4,6)S(4,6) and R(12,6)R(12,6) are equal: ∣PS∣=∣PR∣|P S|=|P R|.

Studdy Solution

STEP 1

Assumptions1. (4,6)(4,6) and R(12,6)R(12,6) are fixed points in the Cartesian plane. . A point (x,y)(x, y) moves in the plane such that the distance from to to is equal to the distance from to $R$.
3. We are asked to find the equation of the locus of , which is the set of all points (x,y)(x, y) that satisfy the given condition.

STEP 2

The distance between two points (x1,y1)(x1, y1) and (x2,y2)(x2, y2) in the Cartesian plane is given by the formulad=(x2−x1)2+(y2−y1)2d = \sqrt{(x2-x1)^2 + (y2-y1)^2}

STEP 3

We are given that the distance from to to is equal to the distance from to $R$. We can write this as||=| R|$$

STEP 4

Substitute the coordinates of , $R$, and into the distance formula(x−4)2+(y−6)2=(x−12)2+(y−6)2\sqrt{(x-4)^2 + (y-6)^2} = \sqrt{(x-12)^2 + (y-6)^2}

STEP 5

Square both sides of the equation to eliminate the square roots(x−4)2+(y−)2=(x−12)2+(y−)2(x-4)^2 + (y-)^2 = (x-12)^2 + (y-)^2

STEP 6

Expand the squares on both sidesx2−8x+16+y2−12y+36=x2−24x+144+y2−12y+36x^2 -8x +16 + y^2 -12y +36 = x^2 -24x +144 + y^2 -12y +36

STEP 7

Notice that y2−12y+36y^2 -12y +36 appears on both sides of the equation, so we can subtract it from both sides to simplifyx2−x+16=x2−24x+144x^2 -x +16 = x^2 -24x +144

STEP 8

Subtract x2x^2 from both sides−8x+16=−24x+144-8x +16 = -24x +144

STEP 9

Add 24x24x to both sides16x+16=14416x +16 =144

STEP 10

Subtract 1616 from both sides16x=12816x =128

STEP 11

Divide both sides by 1616x=8x =8So, the locus of $$ is a vertical line with equation $x =8$.

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