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In the diagram, $PQ_1$, $PQ_2$, and $PQ$ are parallel and so also are $Q_1P_2$ and $Q_2P_3$ - Leaving Cert Mathematics - Question 6B - 2011

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Question 6B

In-the-diagram,-$PQ_1$,-$PQ_2$,-and-$PQ$-are-parallel-and-so-also-are-$Q_1P_2$-and-$Q_2P_3$-Leaving Cert Mathematics-Question 6B-2011.png

In the diagram, $PQ_1$, $PQ_2$, and $PQ$ are parallel and so also are $Q_1P_2$ and $Q_2P_3$. Prove that $|Q_2|\times|P_3|=|P_2|\times|Q_1|.$

Worked Solution & Example Answer:In the diagram, $PQ_1$, $PQ_2$, and $PQ$ are parallel and so also are $Q_1P_2$ and $Q_2P_3$ - Leaving Cert Mathematics - Question 6B - 2011

Step 1

Prove that $|P_3|/|P_1| = |Q_2|/|Q_1|$

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Answer

Given that PQ1∣∣PQ2PQ_1 || PQ_2, by the properties of parallel lines and transversal, the ratios of the segments formed are equal. Thus, we can state that:

∣P3∣∣P1∣=∣Q2∣∣Q1∣\frac{|P_3|}{|P_1|} = \frac{|Q_2|}{|Q_1|}

Step 2

Prove that $|P_2|/|P_1| = |Q_2|/|Q_1|$

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Answer

Similarly, since PQ∣∣PQ2PQ || PQ_2 and with Q1P2Q_1P_2 being another transversal, we imply:

∣P2∣∣P1∣=∣Q2∣∣Q1∣\frac{|P_2|}{|P_1|} = \frac{|Q_2|}{|Q_1|}

Step 3

Using the results to find $|Q_2|\times|P_3|$ and $|P_2|\times|Q_1|$

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Answer

We have:

  1. From the first step: ∣P3∣=∣P1∣×∣Q2∣∣Q1∣|P_3| = |P_1| \times \frac{|Q_2|}{|Q_1|} So therefore: ∣Q2∣×∣P3∣=∣Q2∣×(∣P1∣×∣Q2∣∣Q1∣)=∣Q2∣2∣Q1∣∣P1∣|Q_2| \times |P_3| = |Q_2| \times \left(|P_1| \times \frac{|Q_2|}{|Q_1|}\right) = \frac{|Q_2|^2}{|Q_1|}|P_1|

  2. From the second step: ∣P2∣=∣P1∣×∣Q2∣∣Q1∣|P_2| = |P_1| \times \frac{|Q_2|}{|Q_1|} Hence: ∣P2∣×∣Q1∣=∣Q2∣×∣P1∣|P_2| \times |Q_1| = |Q_2| \times |P_1|

Combining these results, we arrive at:

∣Q2∣×∣P3∣=∣P2∣×∣Q1∣|Q_2| \times |P_3| = |P_2| \times |Q_1|

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