Answered

Welcome to Westonci.ca, the place where your questions find answers from a community of knowledgeable experts. Our platform provides a seamless experience for finding reliable answers from a knowledgeable network of professionals. Join our platform to connect with experts ready to provide precise answers to your questions in different areas.

find the point on the y-axis that is equidistant from (6,1) (-2,-3)

Sagot :

Looking the centre of the circle on the y-axis passing through the points (6;1) and  (-2,-3).
The coordinates of the center of this a circle are equal y-intercept of line segment bisector.

[tex](x-6)^2+(y-1)^2=(x+2)^2+(y+3)^2\\\\x^2-12x+36+y^2-2y+1=x^2+4x+4+y^2+6y+9\\-12x-2y+37=4x+6y+13\\-2y-6y=4x+12x+13-37\\-8y=16x-24\ \ \ \ \ |:(-8)\\y=-2x+3\\\\Answer:(0;\ 3)[/tex]

View image Аноним
Lilith
[tex] Any \ point \ on \ the \ y axis \ can \ be \ stated \ as \ C(0,y) \\ \\ Distance \ Formula:\\\\ Given \ the \ two \ points \ (x _{1}, y _{1}) and (x _{2}, y _{2}), \\ \\the \ distance \ between \ these \ points \ is \ given \ by \ the \ formula: \\ \\ d= \sqrt{(x_{2}-x_{1})^2 +(y_{2}-y_{1})^2} \\ \\CA= \sqrt{(-2-0)^2 +(-3-y)^2}=\sqrt{(-2 )^2 +(-3-y)^2}=\sqrt{4 +9+6y+y^2 }=\\\\-\sqrt{ y^2+6y+13 } [/tex]

[tex]CB= \sqrt{(1-y)^2 +(6-0)^2} =\sqrt{ 1-2y +y^2 +36} =\sqrt{ y^2-2y +37} \\ \\CA = CB \\ \\\sqrt{ y^2+6y+13 } =\sqrt{ y^2-2y +37} \ \ |^2\\\\ y^2+6y+13 = y^2-2y +37 \\ \\y^2+6y- y^2+2y =37-13\\ \\8y = 24 \ \ / :8 \\ \\y= 3 \\ \\C=(0,3)[/tex]
View image Lilith
We appreciate your time. Please come back anytime for the latest information and answers to your questions. Thank you for your visit. We're committed to providing you with the best information available. Return anytime for more. Discover more at Westonci.ca. Return for the latest expert answers and updates on various topics.