Answered

Get detailed and accurate responses to your questions with IDNLearn.com. Our community is here to provide detailed and trustworthy answers to any questions you may have.

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
Lilithidn
[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 participation in this forum. Keep exploring, asking questions, and sharing your insights with the community. Together, we can find the best solutions. Trust IDNLearn.com for all your queries. We appreciate your visit and hope to assist you again soon.