Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

A line passing through two distinct points $P_1(x_1,y_1),P_2(x_2,y_2)$ can be expressed by $$\det\left| \begin{array}{ccc} x-x_1&y-y_1 \\ x_2-x_1&y_2-y_1 \\ \end{array} \right|=0$$

Since line is set $A=\{a P_1+bP_2|a+b=1\}$ and it corresponds to determinant properties.

A circle passing three(non-colinear) points $(x_1,y_1),(x_2,y_2),(x_3,y_3)$ can be expressed by $$\det\left| \begin{array}{ccc} (x-x_1)^2+(y-y_1)^2 & (x-x_1) & (y-y_1) \\ (x_2-x_1)^2+(y_2-y_1)^2 & (x_2-x_1) & (y_2-y_1) \\ (x_3-x_1)^2+(y_3-y_1)^2 & (x_3-x_1) & (y_3-y_1) \end{array} \right|=0$$ or $$\det\left| \begin{array}{ccc} x^2+y^2 & x & y&1 \\ x_1^2+y_1^2 & x_1 & y_1&1 \\ x_2^2+y_2^2 & x_2 & y_2&1 \\ x_3^2+y_3^2 & x_3 & y_3&1 \\ \end{array} \right|=0$$

Please explain simple and neat as possible(like in determinant sense).

share|improve this question
    
For the first one, when the determinant is $0$, that means the first row vector is parallel to the surface formed by the second and third row vectors. It means that $v_1=t v_2+(1-t)v_3$ for $(0\leq t\leq 1)$, where $v_i$ indicates the $i$-th row vector. Or just use David's idea mentioned below. –  Aran Komatsuzaki May 25 at 7:17

2 Answers 2

up vote 2 down vote accepted

Consier the set of equations $$Ax^2+Ay^2+Bx+Cy+D=0 \tag{1} $$ $$Ax_1^2+Ay_1^2+Bx_1+Cy_1+D=0 \tag{2} $$ $$Ax_2^2+Ay_2^2+Bx_2+Cy_2+D=0 \tag{3} $$ $$Ax_3^2+Ay_3^2+Bx_3+Cy_3+D=0 \tag{4} $$ this is a $4\times4$ linear system and it must satisfy, $$\left( \begin{array}{ccc} x^2+y^2 & x & y&1 \\ x_1^2+y_1^2 & x_1 & y_1&1 \\ x_2^2+y_2^2 & x_2 & y_2&1 \\ x_3^2+y_3^2 & x_3 & y_3&1 \\ \end{array} \right) \left( \begin{array}{ccc} A\\ B \\ C \\ D \\ \end{array} \right)=0 \tag{5}$$ taking determinant you get your required condition. Also you can subtract the set of equations $1\to 4$ to reduce it into $3\times 3$ system and get the other result.

share|improve this answer

Hint for the $4\times4$ version.

  • Explain why expanding the determinant gives an equation of the form $$Ax^2+Ay^2+Bx+Cy+D=0\ ,$$ where $A,B,C,D$ are constants.
  • Explain why $A\ne0$.
  • Explain why the three given points satisfy the equation.

Good luck!

share|improve this answer
    
Wow!!!! I have never got two upvotes and two downvotes so quickly! Would anyone care to explain? –  David May 25 at 7:30
    
i'll make it $4-2$ then –  Santosh Linkha May 25 at 7:35

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.