A nice formula for area of the triangle

Asked Oct 02 '15 at 14:36

Active Oct 02 '15 at 14:36

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The following problem can be found in S.L. Loney's The Elements of Coordinate Geometry (Examples X, Problem 19):

Prove that the if the area of the triangle formed by the three straight lines $a_1x+b_1y+c_1=0, a_2x+b_2y+c_2=0, a_3x+b_3y+c_3=0$ is $A$, then

$\displaystyle (a_2b_1-b_2a_1)(a_3b_1-b_3a_1)(a_2b_3-b_2a_3)A = \frac12 \begin{vmatrix} a_1 & b_1 & c_1\\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \\ \end{vmatrix}^2$.

Now one can solve this problem by finding the points of intersections and using the the determinant formula for the area of the triangle.

I want to know if there is a slick proof (especially using vectors). The determinant and the expressions $a_2b_1-b_1a_2$ are reminiscent of cross products.

Thanks.

asked Oct 02 '15 at 14:36

Isomorphism

5,960

7

This question was one of my first on this site. I found Blue's proof really interesting btw. – najayaz Oct 02 '15 at 18:10
@Isomorphism yes, there is a proof by vectors, look at my answer in the duplicated question which uses homogeneous coordinates. – achille hui Nov 05 '16 at 02:49

A nice formula for area of the triangle

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