A moving point is always equidistant from (5,3) and the line 3x+y+5=0. What is the equation of its locus?

Please help. how to do this problem?

the distance of the point (h,k) from the line ax+by+c=0 is

d = |ah+bk+c|/√(a^2+b^2)

the distance between two points (x1, y1) and (x2, y2) is √[ (x2 - x1)^2 + (y2 - y1)^2]

Let the moving point be (x,y). Substituting the values we have:

|3x+5y+5|/√10 = √[(x-5)^2 + (y-3)^2]
(3x+5y+5)^2 = 10((x-5)^2 + (y-3)^2)
9x^2+30xy+25y^2+30x+50y+25 = 10x^2+10y^2-100x-60y+340

-x^2 + 30xy + 15y^2 + 130x + 110y - 315 = 0

Hmm. That's an hyperbola. I was expecting a parabola. Better check my algebra.

why did you put square root of 10 in the denmntor? the formula is Ax1+by1+c/+squareroot of A^2+B^2 SUBSTITUTING THE X AND Y IT WILL YOU A REAL NUMBER

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To solve the problem, we need to find the equation of the locus of the moving point. The locus is the path traced by the moving point as it satisfies the given condition of being equidistant from a fixed point and a line.

Let's break down the steps to solve the problem:

Step 1: Find the coordinates of the fixed point.
Given that the fixed point is (5, 3).

Step 2: Find the distance between the moving point and the fixed point.
Let the coordinates of the moving point be (x, y). The distance between two points (x1, y1) and (x2, y2) is given by the formula:
Distance = √[(x2 - x1)^2 + (y2 - y1)^2]

Using this formula, the distance between the moving point (x, y) and the fixed point (5, 3) is:
Distance = √[(x - 5)^2 + (y - 3)^2]

Step 3: Find the perpendicular distance between the moving point and the given line.
To find the distance between a point (x1, y1) and a line Ax + By + C = 0, the formula for the perpendicular distance is:
Distance = |Ax1 + By1 + C| / √(A^2 + B^2)

In this case, the given line is 3x + y + 5 = 0. So, A = 3, B = 1, and C = 5.
Using the formula, the perpendicular distance between the moving point (x, y) and the line is:
Distance = |3x + y + 5| / √(3^2 + 1^2) = |3x + y + 5| / √10

Step 4: Set the distances equal to each other.
Since the point is equidistant from the fixed point and the line, we can set the two distances equal to each other:
√[(x - 5)^2 + (y - 3)^2] = |3x + y + 5| / √10

Step 5: Square both sides of the equation to eliminate the square root.
Square both sides of the equation:
[(x - 5)^2 + (y - 3)^2] = ((3x + y + 5)^2) / 10

Expand and simplify the equation:
(x^2 - 10x + 25) + (y^2 - 6y + 9) = (9x^2 + 6xy + 10x + 6xy + y^2 + 10y + 5) / 10

Simplify further:
10x^2 - 99x + 10y^2 - 4y + 40 = 0

Therefore, the equation of the locus of the moving point is:
10x^2 - 99x + 10y^2 - 4y + 40 = 0