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I am unable to get the answer to this question. The question is to find the area of a quadrilateral having its vertices as coordinates in order:

$A(3,-2)$; $B(4,0)$; $C(6,-3)$ and $D(5,-5)$.

I divided this into 2 triangles - $\Delta ABC$ and $\Delta BCD$. Now by the area of triangle formula, I am getting the total area to be $35$ square units but my tuition partner is getting a very different answer $7$ square units to be precise.

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closed as off-topic by TastyRomeo, Alex Mathers, user91500, Claude Leibovici, projectilemotion Feb 28 '17 at 12:12

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enter image description here

you need to know determinants for using this. this is more easier and faster way to do it.

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  • $\begingroup$ @NeerjaSharma did you get it? $\endgroup$ – ATHARVA Feb 27 '17 at 15:19
  • $\begingroup$ I couldn't get the precise answer but i am getting 35 sq units is that correct atharv singh patlan? $\endgroup$ – Neerja Sharma Feb 28 '17 at 16:18
  • $\begingroup$ @NeerjaSharma The answer is 7 units BTW what is singh patlan in your comment. $\endgroup$ – ATHARVA Feb 28 '17 at 16:34
  • $\begingroup$ Looks like your friend is correct @NeerjaSharma $\endgroup$ – ATHARVA Feb 28 '17 at 16:36
  • $\begingroup$ @NeerjaSharma Which method did you use this or other one posted by me? $\endgroup$ – ATHARVA Feb 28 '17 at 16:37
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Enclose the quadrilateral in a rectangle $R$ with vertices of $(3,-5),(6,-5),(6,0),(3,0)$. This rectangle will have area $3\cdot 5 =15$ but will overshoot the quadrilateral in exactly four triangular areas.

Finding the area of these triangles should be straight forward, as their bases and heights are parallel the coordinate axes - I found two triangles of area $3$ and another two of area $1$, so all four together will have combined area $2(3)+2(1)=8$.

Take the difference:$15-8=7$.

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  • $\begingroup$ What answer do you get? $\endgroup$ – mrnovice Feb 27 '17 at 15:32
  • $\begingroup$ 35 sq units is it coming out correct? $\endgroup$ – Neerja Sharma Feb 28 '17 at 16:18
  • $\begingroup$ I get an answer of 7 $\endgroup$ – joeb Feb 28 '17 at 18:57
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join diagonal AC. your quadrilateral will get divided into two triangles ABC ADB. find length of AC AB BC CD DA. and use herons formula to find area of two triangles. addition will give you the require area.

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  • $\begingroup$ you can do this in another way i will post that method too in a minute. $\endgroup$ – ATHARVA Feb 27 '17 at 15:13
  • $\begingroup$ @NeerjaSharma did you get it? $\endgroup$ – ATHARVA Feb 27 '17 at 15:19
  • $\begingroup$ I got the answer as 35 sq units but my friend is getting 7 sq units $\endgroup$ – Neerja Sharma Feb 28 '17 at 16:15
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Hint 1

The area of quadrilateral $ABCD$ is equal to the sum of areas of triangles $ABC$ and $ACD$

Hint 2

$$P_{\triangle XYZ} = \frac{|\vec{XY}\times \vec{XZ}|}{2} $$

Where $\times $ is the cross product of vectors: $(a,b)\times (x,y) = (0,0,ay-bx)$

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    $\begingroup$ Shouldn't that be the cross product of the vectors? $\endgroup$ – PM. Feb 27 '17 at 15:17
  • $\begingroup$ Yes, it should. $\endgroup$ – Jaroslaw Matlak Feb 27 '17 at 15:19
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Here's a graph so you can see what is going on visually.

enter image description here

By looking at the graph, or explicitly finding the equations of the lines $BC$, $AD$, $CD$, $AB$ we can see that $AB$ is parallel to $AD$ and $CD$. is parallel to $AB$.

Note that also $|BC| = |CD|$ and $|AB|=|CD|$.

Then we can deduce this quadrilateral is a parallelogram. The area is then equal to the product of the base and its height. That is the product of $AD$ and the perpendicular distance between $AD$ and $BC$.The equation of the line $AD$ is given by $y =-\frac{3}{2}x + \frac{5}{2}$

The gradient equation of the normal to this line is then given by $\frac{-1}{-\frac{3}{2}} = \frac{2}{3}$

This normal starts has a point on $AD$ and a corresponding point on $BC$

The equation of the normal is then given by $y = \frac{2}{3} x -4$

The equation of the line $BC$ is given by $y= -\frac{3}{2}x + 6$

Set these two equations equal to find the point of intersection:

$-\frac{3}{2}x + 6 = \frac{2}{3}x -4 \Rightarrow x =\frac{60}{13}, y = -\frac{12}{13}$

Let's call this point of intersection on $BC$ point $ E = (\frac{60}{13},-\frac{12}{13})$.

Then we have $|AE| = \sqrt{(\frac{60}{13} - 3)^{2} + (\frac{12}{13} - - 2)^{2}} = \sqrt{\frac{145}{13}}$

$|AD| = \sqrt{(3-5)^{2}+(-2+5)^{2}} = \sqrt{13}$

Then the area of $ABCD = \sqrt{\frac{145}{13}} \cdot \sqrt{13} = \sqrt{145}$

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  • $\begingroup$ Mine is coming 35 sq units $\endgroup$ – Neerja Sharma Feb 28 '17 at 16:15

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