If the vertices of a triangle are (am_1^2, 2a | vedclass

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(C) The area of a triangle with vertices $(x_1, y_1), (x_2, y_2), (x_3, y_3)$ is given by $\Delta = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3

Vedclass · Accepted Answer

$a^2| (m_1 - m_2)(m_2 - m_3)(m_3 - m_1) |$

Vedclass · Answer

(C) The area of a triangle with vertices $(x_1, y_1), (x_2, y_2), (x_3, y_3)$ is given by $\Delta = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$.Substituting the given vertices:$\Delta = \frac{1}{2} |am_1^2(2am_2 - 2am_3) + am_2^2(2am_3 - 2am_1) + am_3^2(2am_1 - 2am_2)|$$\Delta = \frac{1}{2} \cdot 2a^2 |m_1^2(m_2 - m_3) + m_2^2(m_3 - m_1) + m_3^2(m_1 - m_2)|$Using the identity $m_1^2(m_2 - m_3) + m_2^2(m_3 - m_1) + m_3^2(m_1 - m_2) = -(m_1 - m_2)(m_2 - m_3)(m_3 - m_1)$,$\Delta = a^2 |-(m_1 - m_2)(m_2 - m_3)(m_3 - m_1)| = a^2 |(m_1 - m_2)(m_2 - m_3)(m_3 - m_1)|$.

List-$I$	List-$II$
$(A)$ Focus	$(I)$ $(4,2)$
$(B)$ Vertex	$(II)$ $(3,2)$
$(C)$ One end of the latus rectum	$(III)$ $(2,3)$
$(D)$ Point of intersection of the axis and directrix	$(IV)$ $(2,4)$
	$(V)$ $(2,2)$

If the vertices of a triangle are $(am_1^2, 2am_1), (am_2^2, 2am_2),$ and $(am_3^2, 2am_3),$ then the area of the triangle is

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