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Let ABCD be a quadrilateral, let E be the intersection of lines AB and CD, and let F be the intersection of lines BC and AD. If the lines AC and BD of this quadrilateral are perpendicular and the slopes of the equations of its sides AB, BC, CD, DA are, respectively, $-1, 2, 3, 4$, which is the slope of equation of line EF?

Besides using the obvious relation between the slopes of lines AC e BD, what do you do?

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  • $\begingroup$ Is such a quadrilateral even possible? $\endgroup$ – Jens Oct 10 '18 at 16:44
  • $\begingroup$ Yes, it´s possible. First, note that the question doesn't say that the quadrilateral is or must be a convex one, thus it can be a concave quadrilateral. Second, this quadrilateral would be outright impossible only if the slopes of two consecutive sides were equal. $\endgroup$ – MrDudulex Oct 10 '18 at 17:05
  • $\begingroup$ Yes, but you require the diagonals to be perpendicular. Do you have an example of such a quadrilateral? $\endgroup$ – Jens Oct 10 '18 at 17:19
  • $\begingroup$ If we are given only the slopes of the equations of the sides of a quadrilateral, we have infinite possible pairs of slopes of the equations of the diagonals. $\endgroup$ – MrDudulex Oct 10 '18 at 18:31
  • $\begingroup$ So when you say the diagonals are perpendicular, do you mean the extended lines of the diagonals are perpendicular? I.e. the diagonals don't necessarily intersect? $\endgroup$ – Jens Oct 10 '18 at 18:58
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The quadrilateral looks concave. Without loss of generality, assume that $D$ lies on the origin of the $xy$-coordinate system, hence, $D=(0,0)$.

Since $m_{AD}=4$ and $m_{CD}=3$, then lines $AD$ and $CD$ should have the following equations: $$AD\implies f(x):=4x\\ CD\implies g(x):=3x$$ Thus, we know that $A=(a,f(a))$ and $C=(c,g(c))$ for $a,c \in \mathbb R$. Since the diagonals of $ABCD$ are perpendicular, then we know that:

$$m_{AC}=-\frac1{m_{DB}}$$

Since we already know points $A,C$, then $m_{AC}$ is: $$m_{AC}=\frac{g(c)-f(a)}{c-a}\implies m_{DB}=-\frac{c-a}{g(c)-f(a)}\\ DB\implies h(x):=-\frac{c-a}{g(c)-f(a)}x\\ \implies B=(b,h(b))\,\,\,\,\forall b \in \mathbb R$$

Now, we have to solve the following equations for $a,c$ in terms of $b$ in order to satisfy the slope requirements: $$m_{AB}=-1=\frac{h(b)-f(a)}{b-a}\\ m_{BC}=2=\frac{h(b)-g(c)}{b-c}$$ And we get one of the solutions as: $$\left\{a\to \frac{1}{55} \left(\sqrt{346} b+26 b\right),c\to \frac{1}{11} \left(\sqrt{346} b-7 b\right)\right\}\tag{1}$$


Now, we can write the following equations of the sides of the quadrilateral: $$AB\implies y=\frac{f\left(a\right)-h\left(b\right)}{a-b}\left(x-b\right)+h\left(b\right)\\ BC\implies y=\frac{g\left(c\right)-h\left(b\right)}{c-b}\left(x-b\right)+h\left(b\right)\\ CD\implies y=3x\\ AD\implies y=4x$$

Solving for the intersection of $AB$ and $CD$ to get $E$, and $BC$ and $AD$ to get $F$, we find that $F$ and $E$ are: $$E=\left(\frac{a b (17 a-13 c)}{4 a^2+13 a b-3 a c-10 b c},\frac{3 a b (17 a-13 c)}{4 a^2+13 a b-3 a c-10 b c}\right)\\ F=\left(\frac{b c (13 a-10 c)}{17 a b-4 a c-13 b c+3 c^2},\frac{4 b c (13 a-10 c)}{17 a b-4 a c-13 b c+3 c^2}\right)\tag{2}$$

Now, using $(1)$ above and substituting it in $(2)$, we get simpler definitions for $E,F$, to wit (again, this should be for all $b\in\mathbb R$): $$E=\left(\frac{1}{44} \left(\sqrt{346}+26\right) b,\frac{3}{44} \left(\sqrt{346}+26\right) b\right)\\ F=\left(\frac{1}{22} \left(\sqrt{346}-7\right) b,\frac{2}{11} \left(\sqrt{346}-7\right) b\right)$$

And thus, we get the slope of $EF$, which should be invariant to $b$, as: $$m_{EF}=\frac{\frac{3}{44} \left(\left(\sqrt{346}+26\right) b\right)-\frac{4}{22} \left(\left(\sqrt{346}-7\right) b\right)}{\frac{1}{44} \left(\sqrt{346}+26\right) b-\frac{1}{22} \left(\sqrt{346}-7\right) b}$$

Which simplifies to: $$\bbox[10px, border:2px solid red]{\therefore m_{EF}=\frac{1}{19} \left(55-\sqrt{346}\right)}\tag3$$


You can check this Desmos implementation. You can check that all the slope requirements were met, and $(3)$ should give you the correct answer.

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  • $\begingroup$ Congratulations, John Glenn! But haven't you missed the other solution? $\endgroup$ – MrDudulex Oct 12 '18 at 18:55
  • $\begingroup$ Btw, this quadrilateral must be concave. $\endgroup$ – MrDudulex Oct 12 '18 at 19:04
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There's a simplier way to solve this kind of problem:

First use the following forgotten theorem of analytic geometry:

If the slopes of the equations of sides AB, BC, CD, DA, of a quadrilateral ABCD are respectively $m_1, m_2, m_3, m_4$, and the diagonals AC and BD (or the extended lines of them) are perpendicular, then the equation in $p$

$$(m_1m_3-m_2m_4)p^2 -((m_2+m_4)(1+m_1m_3)-(m_1+m_3)(1+m_2m_4))p -(m_1m_3-m_2m_4) =0$$

is satisfied by both slopes of the diagonals (in case both exist) or by the only existing one.

Applying this theorem, we get the second degree equation

$$11p^2-30p-11=0$$

Solving it, we get the solutions $${15+\sqrt {346} \over 11}, {15-\sqrt {346} \over 11}$$

for the slopes of the diagonals AC and BD.

Afterwards we use a neat and straightforward formula for calculating the slope of EF (same notation used in the above theorem):

$$m_{EF}={m_2m_3(m_1+m_4-m_{BD})-m_1m_4(m_2+m_3-m_{BD})\over m_2m_3-m_1m_4+(m_1+m_4-m_2-m_3)m_{BD}}$$

And we finally arrive at the solutions:

$$m_{EF}={55+\sqrt{346}\over 19}$$

or $$m_{EF}={55-\sqrt{346}\over 19}$$

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