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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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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
@J.M.: Yes, but it's rather the way the question is asked that annoys me. – Dario Sep 19 '10 at 15:19
@Dario: Oh, then I'm in complete agreement, then. :) – J. M. Sep 19 '10 at 17:54

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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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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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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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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