# Find the point (x, y) equidistant from (1, 2), (5, 8) and (4, 3)?

*print*Print*list*Cite

### 2 Answers

The point equidistant from (1, 2), (4, 3) and (5, 8) is the point of intersection of perpendicular bisectors between any two sets of points.

For the points, (1, 2) and (4, 3) the midpoint is (5/2, 5/2). The slope of the line between them is (3 – 2)/ (4 - 1) = 1 / 3. The slope of the perpendicular line is the negative reciprocal or -3. As the line passes through (5/2, 5/2), its equation is (y – 5/2) = -3*(x - 5/2)

=> (y – 5/2) = -3x + 15/2

=>y + 3x– 5/2 - 15/2 = 0

=> 3x + y – 10 = 0

For the points, (4, 3) and (5, 8) the midpoint is (9/2, 11/2). The slope of the line between them is (8 – 3)/ (5 - 4) = 5 / 1 = 5. The slope of the perpendicular line is the negative reciprocal or -1/5. As the line passes through (9/2, 11/2), its equation is (y – 11/2) = (-1/5)*(x - 9/2)

=> 5*(y – 11/2) = - x + 9/2

=> 5y – 55/2 + x – 9/2 = 0

=> x + 5y – 32 = 0

The point of intersection of 3x + y – 10 = 0 and x + 5y – 32 = 0 can be determined by:

x + 5y – 32 = 0

=> x = 32 – 5y

Substitute in 3x + y – 10 =0

=> 3(32 – 5y) + y – 10 = 0

=> 96 – 15y + y – 10 = 0

=> 14y = 86

=> y = 43/7

x = 32 – 5*(43/7)

=> x = 9/7

**Therefore the required point is (9/7, 43/7)**

The equidistant point from (1,2), (5,8) and (4,3) is the circum centre of the the circum circle passing through the points (1, 2), (5, 8) and (4, 3).

Let (h,k) be the centre and r be the radius of this circle. Then the equation the circum circle is (x-h)^2+(y-k)^2 = r^2....(A)

(1,2), (5,8) lies on (A):

=> (1-h)^2+(2-k)^2 = r^2 = (5-h)^2+(8-k)^2 . This simplifies to

-2h-4k +5 = -10h-16k+25+64.

8h+12k = 89-5 = 84.

2h +3k = 21.................(1).

Similarly (1,2) and (4,3) are on the circle (A)

=> (1-h)^2+(2-k)^2 = (4-h)^2+(3-k)^2.

=> -2k-4k+5 = -8h-6k+16+9 .

6h+2k = 25 - 5 = 20.

=> 3h+k = 10...(2).

From (2), k = 10-3h. Substituting k = 10-3h in (1), we get:

2h+3(10-3h) = 21.

2h- 9h = 21- 30 = -9.

-7h = -9. Or h = -9/-7 = 9/7.

Substituting the value of h = 9/7 in (2), we get:

3(9/7)+k = 10. So k = 10-3(9/7) = (70-23)/7 = 47/7

Therefore the circum centre of the the circum circle passing through the points (1, 2), (5, 8) and (4, 3) is (9/7, 47/7).