Gujarat BoardEnglish MediumSTD 10MathsCoordinate Geometry3 Marks
Question
Do the points $(3, 2), (–2, –3)$ and $(2, 3)$ form a triangle$?$ If so, name the type of triangle formed.
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Answer
Let us apply the distance formula to find the distances $PQ, QR$ and $PR,$ where
$P \leftrightarrow (3, 2)$,
$Q \leftrightarrow (–2, –3)$ and
$R \leftrightarrow (2, 3)$
are the given points. We have
$\mathrm{PQ}=\sqrt{(3+2)^{2}+(2+3)^{2}}=\sqrt{5^{2}+5^{2}}=\sqrt{50}=7.07(\text { approx. })$
$\mathrm{QR}=\sqrt{(-2-2)^{2}+(-3-3)^{2}}=\sqrt{(-4)^{2}+(-6)^{2}}=\sqrt{52}=7.21(\text { approx. })$
$\mathrm{PR}=\sqrt{(3-2)^{2}+(2-3)^{2}}=\sqrt{1^{2}+(-1)^{2}}=\sqrt{2}=1.41(\text { approx. })$
Since the sum of any two of these distances is greater than the third distance, therefore, the points $P, Q$ and $R$ form a triangle.
Also, $P Q^2+P R^2=Q R^2$, by the converse of Pythagoras theorem, we have $\angle P = 90^\circ$. Therefore, $PQR$ is a right triangle.
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