Question
Show that the given points form a right angled triangle and check whether they satisfy Pythagoras theorem
A(1, – 4), B(2, – 3) and C(4, – 7)
A(1, – 4), B(2, – 3) and C(4, – 7)
✓
Answer
The vertices are $A(1,-4), B(2,-3)$ and $C(4,-7)$
Slope of a line $=\frac{y_2-y_1}{x_2-x_1}$
Slope of $A B=\frac{-3+4}{2-1}=\frac{1}{1}=1$
Slope of $B C=\frac{-7+3}{4-2}=\frac{-4}{2}=-2$
Slope of $AC =\frac{-7+4}{4-1}=-\frac{3}{3}=-1$
Slope of $A B \times$ Slope of $A C=1 \times-1=-1$
$\therefore A B$ is $\perp^r$ to $A C$
$
\angle A =90^{\circ}
$
$\therefore A B C$ 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} \\
& AB =\sqrt{(2-1)^2+(-3+4)^2} \\
& =\sqrt{1^2+1^2} \\
& =\sqrt{2} \\
& BC =\sqrt{(4-2)^2+(-7+3)^2} \\
& =\sqrt{(2)^2+(-4)^2} \\
& =\sqrt{4+16} \\
& =\sqrt{20}
\end{aligned}
$
$
\begin{aligned}
& A C=\sqrt{(4-1)^2+(-7+4)^2} \\
& =\sqrt{3^2+(-3)^2} \\
& =\sqrt{9+9} \\
& =\sqrt{18} \\
& B C^2=A B^2+A C^2 \\
& (\sqrt{20})^2=(\sqrt{2})^2+(\sqrt{18})^2 \\
& 20=2+18 \\
& 20=20
\end{aligned}
$
$\Rightarrow$ Pythagoras theorem verified.
Slope of a line $=\frac{y_2-y_1}{x_2-x_1}$
Slope of $A B=\frac{-3+4}{2-1}=\frac{1}{1}=1$
Slope of $B C=\frac{-7+3}{4-2}=\frac{-4}{2}=-2$
Slope of $AC =\frac{-7+4}{4-1}=-\frac{3}{3}=-1$
Slope of $A B \times$ Slope of $A C=1 \times-1=-1$
$\therefore A B$ is $\perp^r$ to $A C$
$
\angle A =90^{\circ}
$
$\therefore A B C$ 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} \\
& AB =\sqrt{(2-1)^2+(-3+4)^2} \\
& =\sqrt{1^2+1^2} \\
& =\sqrt{2} \\
& BC =\sqrt{(4-2)^2+(-7+3)^2} \\
& =\sqrt{(2)^2+(-4)^2} \\
& =\sqrt{4+16} \\
& =\sqrt{20}
\end{aligned}
$
$
\begin{aligned}
& A C=\sqrt{(4-1)^2+(-7+4)^2} \\
& =\sqrt{3^2+(-3)^2} \\
& =\sqrt{9+9} \\
& =\sqrt{18} \\
& B C^2=A B^2+A C^2 \\
& (\sqrt{20})^2=(\sqrt{2})^2+(\sqrt{18})^2 \\
& 20=2+18 \\
& 20=20
\end{aligned}
$
$\Rightarrow$ Pythagoras theorem verified.
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