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The perpendicular bisector of the line joining $A(0,1)$ and $C(-4,7)$ intersects the $x$-axis at $B$ and the $y$-axis at $D$. Find the area of the quadrilateral.

Thank you in advance!

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@Agusti: consensus on meta was to not use [homework] tag as speculation outside the question as it is posted. I have advocated the creation of tags such as [unsourced] and [generic-question], or whatever else would carry useful but objective information. Here, for example, there is no source for the problem, nor anything special about the points (0,1), (-4,7). Maybe a [numerical], [arithmetic] or (less plausibly) [specialized] tag can become standard for the questions with oddly specific data. – T.. Sep 19 '10 at 4:00
@T.. Thanks. I didn't know what the consensus about this issue was (should take a look at meta more often) and I see your point. I've retagged another question like this: I'll look for it and erase my "homework" tag. – a.r. Sep 19 '10 at 4:17
@Agusti: not necessary to erase anything, there will be a FAQ eventually and those questions are probably defunct. I commented to remind the users of meta that, despite the disagreement from some StackOverflow hard-liners (and as Bill D. mentioned, math is different), having additional tags that are not fields of mathematics could be very useful. It is not reasonable to expect that having just a self-disclosed [homework] tag will address the issues raised by [unsourced], [numerical], [hyperspecific], [computation-request] or other such material (much of which is not homework at all). – T.. Sep 19 '10 at 4:25
So am I wrong, or is site about providing help? This question doesn't ask for help, it just wants us to do someone's work. – Dario Sep 19 '10 at 10:34
@Dario: so far as I can tell none of the answers thus far has given an explicit solution; I suppose nothing here can be construed as spoonfeeding. – J. M. Sep 19 '10 at 14:22

5 Answers 5

The perpendicular bisector will pass through the mid-point: $(-2, 4)$ of the line passing through the points $A(0, 1)$ & $C(-4, 7)$ & have a slope: $\frac{2}{3}$ normal to the line $AC$.

The equation of perpendicular bisector is given as $$y-4=\frac{2}{3}(x-(-2))$$ $$\implies y=\frac{2x+4+12}{3}=\frac{2x+16}{3}$$ By substituting $y=0$ & $x=0$ respectively, the points of intersection of perpendicular bisector with the axes are determined as : $B(-8, 0)$ & $D\left(0, \frac{16}{3}\right)$. Now divide the quadrilateral $ABCD$ into two triangles $\Delta ABC$ & $\Delta ACD$ Thus we have $$\text{area of quadrilateral}\space ABCD=\text{area of}\space \Delta ABC \space \text{with vertices}\space (0, 1),(-8, 0)\space \text{&}\space (-4, 7)+\text{area of }\space \Delta ACD \space \text{with vertices}\space (0, 1),(-4, 7)\space \text{&}\space \left(0, \frac{16}{3}\right)$$ $$\implies \text{area of quadrilateral}\space ABCD=\frac{1}{2}\left|0(0-7)-8(7-1)-4(1-0)\right|+\frac{1}{2}\left|0\left(7-\frac{16}{3}\right)-4\left(\frac{16}{3}-1\right)+0(1-7)\right|$$ $$=\frac{1}{2}\left|-60\right|+\frac{1}{2}\left|\frac{-52}{3}\right|=30+\frac{26}{3}$$$$=\frac{116}{3}\space \text{sq.unit}$$

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Making Weltschmerz's answer explicit, and very much related to Isaac's answer: note that the quadrilateral can be split into two triangles. The area of one triangle is half the product of the length of $\overline{BD}$ (the base) and half the length of $\overline{AC}$ (the height). Doubling that gives an area expression that is exactly what Isaac stated.

As for computing the coordinates of $B$ and $D$, here's a twofer method: once you can compute the equation of the line from the point-slope form, transform the equation you have into the "two-intercept" form


whence your x- and y-intercepts are (a,0) and (0,b).

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If the quadrilateral is ABCD--that is, if AC and BD are its diagonals--then the quadrilateral has perpendicular diagonals, so its area is half the product of the lengths of the diagonals (which you can find from the coordinates of A, B, C, and D).

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The perpendicular bisector of the segment $AC$, by definition, passes through the midpoint of the segment. Look at your drawing. The segment $BD$ is a base to the triangles $BCD$ and $BAD$. Note also that the heights of the triangles are equal.

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Tag the question as homework and show what you've already tried, otherwise it will be (rightfully) closed soon.

Now, we just had this question.


  • find middle of $\overrightarrow{AC}$
  • create a line's equation $b: y = m\cdot x + c$ for the bisector
  • compute the intersections of $b$ and the coordinate axis
  • compute the area
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In 9 hours there was not one vote to close. We need tags that any editor can apply, not only the self-disclosed [homework] tag (which is speculative and subjective to add when the homework status is not disclosed in the question). – T.. Sep 19 '10 at 4:33

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