Question
Show that the given points form a right angled triangle and check whether they satisfy Pythagoras theorem
L(0, 5), M(9, 12) and N(3, 14)
L(0, 5), M(9, 12) and N(3, 14)
✓
Answer
The vertices are $L(0,5), M(9,12)$ and $N(3,14)$
Slope of a line $=\frac{y_2-y_1}{x_2-x_1}$
Slope of LM $=\frac{12-5}{9-0}=\frac{7}{9}$
Slope of $MN =\frac{14-12}{3-9}=\frac{2}{-6}=-\frac{1}{3}$
Slope of $LN =\frac{4-5}{3-0}=\frac{9}{3}=3$
Slope of MN $\times$ Slope of LN $=-\frac{1}{3} \times 3=-1$
$\therefore MN \perp LN$
$
\angle N =90^{\circ}
$
$\therefore LMN$ is a right angle triangle
Verification:
$
\begin{aligned}
& \text { Distance }=\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2} \\
& LN =\sqrt{(3-0)^2+(14-5)^2} \\
& =\sqrt{3^2+9^2} \\
& =\sqrt{9+81} \\
& =\sqrt{90} \\
& =\sqrt{6^2+(-2)^2}=\sqrt{(9-3)^2+(12-14)^2} \\
& =\sqrt{36+4} \\
& =\sqrt{40} \\
& =\sqrt{9^2+7^2}=\sqrt{(9-0)^2+(12-5)^2}
\end{aligned}
$
$
\begin{aligned}
& =\sqrt{9^2+7^2} \\
& =\sqrt{81+49} \\
& =\sqrt{130} \\
& LM ^2= LN ^2+ MN ^2 \\
& 130=90+40 \\
& 130=130
\end{aligned}
$
$\Rightarrow$ Pythagoras theorem is verified
Slope of a line $=\frac{y_2-y_1}{x_2-x_1}$
Slope of LM $=\frac{12-5}{9-0}=\frac{7}{9}$
Slope of $MN =\frac{14-12}{3-9}=\frac{2}{-6}=-\frac{1}{3}$
Slope of $LN =\frac{4-5}{3-0}=\frac{9}{3}=3$
Slope of MN $\times$ Slope of LN $=-\frac{1}{3} \times 3=-1$
$\therefore MN \perp LN$
$
\angle N =90^{\circ}
$
$\therefore LMN$ is a right angle triangle
Verification:
$
\begin{aligned}
& \text { Distance }=\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2} \\
& LN =\sqrt{(3-0)^2+(14-5)^2} \\
& =\sqrt{3^2+9^2} \\
& =\sqrt{9+81} \\
& =\sqrt{90} \\
& =\sqrt{6^2+(-2)^2}=\sqrt{(9-3)^2+(12-14)^2} \\
& =\sqrt{36+4} \\
& =\sqrt{40} \\
& =\sqrt{9^2+7^2}=\sqrt{(9-0)^2+(12-5)^2}
\end{aligned}
$
$
\begin{aligned}
& =\sqrt{9^2+7^2} \\
& =\sqrt{81+49} \\
& =\sqrt{130} \\
& LM ^2= LN ^2+ MN ^2 \\
& 130=90+40 \\
& 130=130
\end{aligned}
$
$\Rightarrow$ Pythagoras theorem is verified
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