Area of a quadrilateral

Question

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!

Answer 1

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

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Now, we just had this question.

Thus

• 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

Answer 2

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.

Answer 3

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).

Answer 4

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.

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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

$\frac{x}{a}+\frac{y}{b}=1$

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