# Find the equation of the locus at a point moving so that it is equidistant from the point (2,2) and the line y=-4.

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### 3 Answers

You should find the equation of the line that is perpendicular to the segment that connects the point `(2,2)` to the point on the line `y=-4` .

You need to find first the equation of the segment passes through the point `(2,2)` and `(4,-4)` such that:

`(4-2)/(x-2) = (-4-2)/(y-2) => 2/(x-2) = (-6)/(y-2)`

`-6(x-2) = 2(y-2) => -3(x-2) = y-2`

You need to convert the equation of the line `-3(x-2) = y- 2` into the slope intercept form such that:

`y = -3x + 6 + 2 => y = -3x + 8`

Hence, the slope of the segment line is `m = -3` .

You need to remember that the product of slopes of two perpendicular lines is -1, hence, you may find the slope of equation of locus points such that:

`m = 1/3`

You need to find the midpoint of segment line whose endpoints are (2,2) and (4,-4) such that:

`x = (2+4)/2 => x= 3`

`y = (2-4)/2 => y = -1`

Hence, you need to find the equation of locus points that passes through the point `(3,-1) ` and it has the slope `m=1/3` such that:

`y + 1 = (1/3)(x - 3) => y = (1/3)x - 1 - 1 => y = (1/3)x - 2`

**Hence, evaluating the equation of locus points under the given conditions yields `y = (1/3)x - 2` .**

### User Comments

Sorry! There is a mistake in the solution posted by me which is corrected hereunder:

The locus of point equidistant from point (2,2) and line y=-4 can be determined as under:

Let the point be (x,y)

Distance of (x,y) from line y=-4 is equal to y-(-4) = y+4

Distance of (x,y) from point (2,2) = sqrt((x-2)^2+(y-2)^2)

therefore:

y+4 = sqrt((x-2)^2+(y-2)^2)

squring both sides we get:

(y+4)^2 = (x-2)^2+(y-2)^2

=> y^2+8y+16 = (x-2)^2+y^2-4y+4

=> 12y = (x-2)^2-12

=> y = (x-2)^2/12-1

**So the locus of the point equidistant from point (2,2) and line y=-4 is given by y = (x-2)^2/12-1 and is a parabola**

The locus of point equidistant from point (2,2) and line y=-4 can be determined as under:

Let the point be (x,y)

Distance of (x,y) from line y=-4 is equal to x-(-4) = x+4

Distance of (x,y) from point (2,2) = sqrt((x-2)^2+(y-2)^2)

therefore:

sqrt((x-2)^2+(y-2)^2) = x+4

squring both sides we get:

(x-2)^2+(y-2)^2 = (x+4)^2

=> x^2-4x+4+(y-2)^2 = x^2+4x+16

=> (y-2)^2 = 8x+12

**So the locus of the point equidistant from point (2,2) and line y=-4 is given by (y-2)^2 = 8x+12**