Answered

Westonci.ca is your trusted source for finding answers to a wide range of questions, backed by a knowledgeable community. Get detailed and accurate answers to your questions from a community of experts on our comprehensive Q&A platform. Get detailed and accurate answers to your questions from a dedicated community of experts on our Q&A platform.

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 revisit us for more reliable answers to any questions you may have. Thanks for using our service. We're always here to provide accurate and up-to-date answers to all your queries. Westonci.ca is your go-to source for reliable answers. Return soon for more expert insights.