Question: In the trapezoid $ABCD, A(4,5), B(1,1), C(-1,4), \overleftrightarrow{AB}\parallel\overleftrightarrow{CD},\overline{CD}=2$, find the coordinate of $D$.
My attempt:
because $\overleftrightarrow{AB}\parallel\overleftrightarrow{CD}$, therefore they have the same slope, we get $$m_{\overleftrightarrow{CD}}=\frac{4}{3}$$By using the point-slope formula, we get the linear equation of $\overleftrightarrow{CD}$ is $$L_{\overleftrightarrow{CD}}:y-4=\frac{4}{3}(x+1)$$which is $$L_{\overleftrightarrow{CD}}:4x-3y=-16$$because $D$ is on $L_{\overleftrightarrow{CD}}$, therefore we can assume the coordinate of $D$ is $$D(t, \frac{4t+16}{3})$$by using $\overline{CD}=2$, sqare both sides we get $$(t+1)^2+(\frac{4t+16}{3}-4)^2=4$$after simplifying and factoring we get$$(5t-1)(5t+11)=0$$we get $t=\frac{1}{5}$ or $-\frac{11}{5}$, therefore the coordinate of $D$ is$$(\frac{1}{5}, \frac{28}{5})\text{or}(-\frac{11}{5}, \frac{12}{5})$$
But there's only one answer $(\frac{1}{5}, \frac{28}{5})$, why? $D$ can be above $C$ or below it, can't it?
$\endgroup$ 11 Answer
$\begingroup$Referring to the graph below, because the other point $D'$ does not make it trapezoid, but a broken line.
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Note: The ordering is important: $ABCD$ and $ABDC$ are different figures.
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