Question

A quadrilateral ABCD is divided by the diagonal AC into two triangles of equal areas. If A, B, C are respectively (3, 4), (-3, 6), (-5, 1), then the locus of D is​

Answer 1

Step-by-step explanation:

this solve will help you

Answer 2

Answer:

Concept:

A quadrilateral is a closed shape and a type of polygon that has four sides, four vertices, and four angles. It is formed by joining four non-collinear points. The sum of interior angles of quadrilaterals is always equal to 360 degrees.

Step-by-step explanation:

Given:

A quadrilateral ABCD is divided by the diagonal AC into two triangles of equal areas.

A, B, C are respectively (3, 4), (-3, 6), (-5, 1)

Find:

Find D

Solution:

let  (x, y) be the fourth vertex of the parallelogram.

                $&A r(\Delta A B C)=A r ( \Delta A C D)$

         $\Delta=\frac{1}{2}\left|x_{1}\left(y_{2}-y_{3}\right)+x_{2}\left(y_{3}-y_{1}\right)+x_{3}\left(y_{1}-y_{2}\right)\right|$

$\frac{1}{2}|3(6-1)-3(1-4)-5(y-6)| =\frac{1}{2} \mid 3(1-y)-5(y-4)+2(4-1) \mid \\ \Rightarrow \mid 3(5)-3(-3)-5(-2||=\mid 3-3 y-5 y+20+3x \\ \Rightarrow|15+9+10|=|3 x-8 y+23| \Rightarrow|3 4|=|3 x-8 y+23| \\ \Rightarrow \mid 3 x-8 y+23=\pm 3 4$

$\begin{array}{ll}3 x-8 y+23=3 y & 3 x-8 y+23=-3 y \\3 x-8 y+23-3 y=0 & 3 x-8 y+23+3 y=0 \\3 x-8 y-11=0 & \Rightarrow 3 x-8 y+57=0\end{array}$

The  locus of D is $(3 x-8 y-11)(3 x-8 y+57)=0$

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