| 1. |
Find the circumcenter of the trianglewhose sides are given by x + y + 2 = 0,5x - y - 2 = 0 and x - 2y + 5 = 0 |
|
Answer» ∆ Start by writing the Equation for circumcircle, so that you can pretty easily determine the center. ∆ Equation for circumcircle is given by : C : \alpha L_1L_2 + \beta L_2L_3 + \gamma L_3L_1 = 0C:αL 1
L 2
+βL 2
L 3
+γL 3
L 1
=0 Note that all three vertices of the ∆ satisfies the above equation. ✓ Your task is to equate the coefficients of squares of x and y and to remove any term of ( xy ) \implies 5 \alpha + 5 \beta + \gamma = - \alpha + 2 \beta - 2 \gamma⟹5α+5β+γ=−α+2β−2γ Additionally, 4 \alpha - 11 \beta - \gamma = 0 \: (removing \: all \: terms \: of \: xy )4α−11β−γ=0(removingalltermsofxy) \begin{gathered}put \: \alpha = 5 \\ \\ \implies \: \beta = 3 , \gamma = - 13\end{gathered} putα=5 ⟹β=3,γ=−13
HENCE, our Equation for Circumcircle is : C : 5 L_1L_2 + 3 L_2L_3 - 13L_3L_1 = 0C:5L 1
L 2
+3L 2
L 3
−13L 3
L 1
=0 And THEREFORE, the center will be : ( - \frac{ 8 \alpha + 23 \beta + 7 \gamma }{2} , \frac{4 \alpha + \beta - \gamma }{2} )(− 2 8α+23β+7γ
, 2 4α+β−γ
) _________________________________________ ✓ That's ONE way of doing it. ✓ You can also try finding the vertices, writing a family through two vertex and making it pass through the third and hence FIND the circumcircle... ∆ But at the end, |
|