Question

Find the coordinate of the point equistant from three point a(5,3) b (5,-6) and c(1,-5)

Answer 1

Answer:

The coordinates of the point which is equidistant from the points A, B and C are

$\sf\:(\:-\:\frac{1}{6}\:,\:-\:\frac{3}{2}\:)$

Step-by-step-explanation:

NOTE: Kindly refer to the attachment first. The figure provided in the attachment is rough figure.

We have given three points which are not collinear.

$\therefore$ By joining those three points, we get $\sf\:\triangle\:ABC$.

$\sf\:A\:\equiv\:(\:5,\:3\:)\:\equiv\:(\:x_{1},\:y_{1}\:)\\\\\sf\:B\:\equiv\:(\:5,\:-6\:)\:\equiv\:(\:x_{2},\:y_{2}\:)\\\\\sf\:C\:\equiv\:(\:1,\:-5\:)\:\equiv\:(\:x_{3},\:y_{3}\:)$

We have to find the point which is equidistant from each of the three vertices of $\sf\:\triangle\:ABC$ i. e. A, B and C.

Let P ( x, y ) be the point which is equidistant from three points A, B and C.

$\therefore\:\sf\:d\:(\:P,\:A\:)\:=\:d\:(\:P, \:B\:)\;=\:d\:(\:P, \:C\:)\\\\\therefore\:\sf\:d\:(\:P, \:A\:)\:=\:d\:(\:P, \:B\:)\\\\\implies\sf\:\sqrt{\:(\:x\:-\:5\:)^{2}\:+\:(\:y\:-\:3\:)^{2}}\:=\:\sqrt{\:(\:x\:-\:5\:)^{2}\:+\:[\:y\:-\:(\:-\:6\:)\:]^{2}}$

By taking square of both sides, we get,

$\therefore\sf\:\cancel{(\:x\:-\:5\:)^{2}}\:+\:(\:y\:-\:3\:)^{2}\:=\:\cancel{(\:x\:-\:5\:)^{2}}\:+\:(\:y\:+\:6\:)^{2}\\\\\implies\sf\:(\:y\:-\:3\:)^{2}\:=\:(\:y\:+\:6\:)^{2}\\\\\implies\sf\:y^{2}\:-\:2\:\times\:y\:\times\:3\:+\:3^{2}\:=\:y^{2}\:+\:2\:\times\:y\:\times\:6\:+\:6^{2}\\\\\implies\sf\:\cancel{y^{2}}\:-\:6y\:+\:9\:=\:\cancel{y^{2}}\:+\:12y\:+\:36\\\\\implies\sf\:-\:6y\:-\:12y\:=\:36\:-\:9\\\\\implies\sf\:-\:18y\:=\:27\\\\\implies\sf\:y\:=\:-\:\frac{\cancel27}{\cancel18}\\\\\implies\boxed{\red{\sf\:y\:=\:-\:\frac{3}{2}}}$

Now,

$\sf\:d\:(\:P, \:B\:) \:=\:d\:(\:P, \:C\:)\\\\\therefore\sf\:\sqrt{\:(\:x\:-\:5\:)^{2}\:+\:[\:y\:-\:(\:-\:6\:)\:]^{2}}\:=\:\sqrt{\:(\:x\:-\:1\:)^{2}\:+\:[\:y\:-\:(\:-\:5\:)\:]^{2}}$

By taking square of both sides, we get,

$\sf\:(\:x\:-\:5\:)^{2}\:+\:(\:y\:+\:6\:)^{2}\:=\:(\:x\:-\:1\:)^{2}\:+\:(\:y\:+\:5\:)^{2}\\\\\implies\sf\:(\:x\:-\:5\:)^{2}\:-\:(\:x\:-\:1\:)^{2}\:=\:(\:y\:+\:5\:)^{2}\:-\:(\:y\:+\:6\:)^{2}\\\\\implies\sf\:\cancel{x^{2}}\:-\:10x\:+\:25\:-\:\cancel{x^{2}}\:-\:2x\:+\:1\:=\:\cancel{y^{2}}\:+\:10y\:+\:25\:-\:\cancel{y^{2}}\:+\:12y\:+\:36$$\\\implies\sf\:-\:10x\:+\:25\:-\:2x\:+\:1\:=\:10y\:+\:25\:+\:12y\:+\:36$$\\\implies\sf\:-\:12x\:+\:26\:=\:22y\:+\:61$$\\\implies\sf\:-\:12x\:=\:22y\:+\:61\:-\:26$$\\\implies\sf\:-\:12x\:=\:22y\:+\:35\:\:-\:-\:(\:1\:)$

Now, by substituting $\sf\:y\:=\:-\:\frac{3}{2}$ in equation ( 1 ), we get,

$\sf\:-\:12x\:=\:22y\:+\:35\:\:\:-\:-\:-\:(\:1\:)\\\\\implies\sf\:-\:12x\:=\:\cancel{22}\:\times\:-\:\frac{3}{\cancel2}\:+\:35\\\\\implies\sf\:-\:12x\:=\:11\:\times\:(\:-\:3\:)\:+\:35\\\\\implies\sf\:-\:12x\:=\:-\:33\:+\:35\\\\\implies\sf\:-\:12x\:=\:2\\\\\implies\sf\:x\:=\:-\:\frac{\cancel2}{\cancel12}\\\\\implies\boxed{\red{\sf\:x\:=\:-\:\frac{1}{6}}}$

Additional Information:

Formula to find the distance between two points:

$\boxed{\red{\sf\:d\:(\:A,\:B\:)\:=\:\sqrt{\:(\:x_{1}\:-\:x_{2}\:)^{2}\:+\:(\:y_{1}\:-\:y_{2}\:)^{2}}}}$