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Question: If the area of a triangle formed by the points $(2a,b), (a + b, 2b + a), (2b,2a)$ is $\lambda$, then what is the area of the triangle whose vertices are $(a+b,a-b),(3b-a,b+3a) and (3a-b,3b-a)$ ?

I attempted to find a relation between the two sets of vertices, but was unable to do so. How exactly would I proceed with this question?

Gummy bears
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2 Answers2

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If you are familiar with determinants then the area of a triangle formed by points with vertices $(x_i,y_i)$ with $i=1,2,3$ is given by $$ \text{Area} = \frac{1}{2}\left| \begin{array}{ccc} 1 & x_1 & y_1\\ 1 & x_2 & y_2\\ 1 & x_3 & y_3\\ \end{array} \right| $$ Using this you can express the area of the first triangle as well as the second triangle. Then use the properties of determinants to get a relation between the two.

Note: The area of the second will be $4$ times that of the first.

Anurag A
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Let $P=(2a,b), \;Q=(a+b,2b+a), \;R=(2b,2a)$ and

$u=\vec{PQ}=\langle b-a,b+a\rangle$ and $v=\vec{QR}=\langle b-a, a-2b\rangle$.

Similarly, let $P_1=(a+b,a-b), \;Q_1=(3b-a,b+3a), \;R_1=(3a-b,3b-a)$ and

$u_1=\vec{P_1Q_1}=\langle 2b-2a, 2b+2a\rangle$ and $v_1=\vec{P_1R_1}=\langle 2a-2b, 4b-2a\rangle$.

Then $A_1=\frac{1}{2}||u\times v||=\lambda$ and $A_2=\frac{1}{2}||u_1\times v_1||=4\lambda$ since $u_1=2u$ and $v_1=-2v$.

user84413
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