Question
Let A(3, – 4), B(9, – 4), C(5, – 7) and D(7, – 7). Show that ABCD is a trapezium.
✓
Answer
Let A(3, – 4), B(9, – 4), C(5, – 7) and D(7, – 7) are the vertices of a quadrilateral.
Slope of a line $=\frac{y_2-y_1}{x_2-x_1}$
Slope of $A B=\frac{-4+4}{9-3}=\frac{0}{6}=0$
Slope of $BC =\frac{-7+4}{5-9}=\frac{-3}{-4}=\frac{3}{4}$
Slope of $C D=\frac{-7+7}{7-5}=\frac{0}{2}=0$
Slope of $AD =\frac{-7+4}{7-3}=\frac{-3}{4}=-\frac{3}{4}$
The slope of $A B$ and $C D$ are equal.
$\therefore A B$ is parallel to $C D$. Similarly, the slope of $A D$ and $B C$ are not equal.
$\therefore A D$ and $B C$ are not parallel.
$\therefore$ The Quadrilateral $A B C D$ is a trapezium.
Slope of a line $=\frac{y_2-y_1}{x_2-x_1}$
Slope of $A B=\frac{-4+4}{9-3}=\frac{0}{6}=0$
Slope of $BC =\frac{-7+4}{5-9}=\frac{-3}{-4}=\frac{3}{4}$
Slope of $C D=\frac{-7+7}{7-5}=\frac{0}{2}=0$
Slope of $AD =\frac{-7+4}{7-3}=\frac{-3}{4}=-\frac{3}{4}$
The slope of $A B$ and $C D$ are equal.
$\therefore A B$ is parallel to $C D$. Similarly, the slope of $A D$ and $B C$ are not equal.
$\therefore A D$ and $B C$ are not parallel.
$\therefore$ The Quadrilateral $A B C D$ is a trapezium.
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