If `(x_i,y_i),i=1,2,3` are the vertices of an equilateral triangle such that `(x_1+2)^2+(y_1-3)^2=(x_2+2)^2+(y_2-3)^2=(x_3+2)^2+(y_3-3)^2`, then find

Question

If `(x_i,y_i),i=1,2,3` are the vertices of an equilateral triangle such that `(x_1+2)^2+(y_1-3)^2=(x_2+2)^2+(y_2-3)^2=(x_3+2)^2+(y_3-3)^2`, then find

If `(x_i,y_i),i=1,2,3` are the vertices of an equilateral triangle such that `(x_1+2)^2+(y_1-3)^2=(x_2+2)^2+(y_2-3)^2=(x_3+2)^2+(y_3-3)^2`, then find the value of `(x_1+x_2+x_3)/(y_1+y_2+y_3)`

Monjur Alahi

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2025-01-04 04:42

1 Answers

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Salim Ahmed Kawsar · Accepted Answer · 2023-02-05T15:29:48+06:00

Correct Answer - `-2//3`
`(x_1+2)^2+(y_1-3)^2=(x_2+2)^2+(y_2-3)^2=(x_3+2)^2+(y_3-3)^2`
Therefore,the circumcenter of the triangle formed by points `(x_1,y_1),(x_2,y_2,(x_3,y_3)` is `(-2,3)`. But the triangle is equilateral, so,that centroid is `(-2,3)`.Therefore,
`(x_1+x_2+x_3)/(3)=-2,(y_1+y_2+y_3)/(3)=3`
or `(x_1+x_2+x_3)/(y_1+y_2+y_3)=-(2)/(3)`