Without using the distance formula,show that the points $(-2,-1), (4,0), (3,3),$ and $(-3,2)$ are the vertices of a parallelogram.
(N/A) Let the points $(-2,-1), (4,0), (3,3),$ and $(-3,2)$ be denoted by $A, B, C,$ and $D$ respectively.
The slope of $AB = \frac{0 - (-1)}{4 - (-2)} = \frac{1}{6}$.
The slope of $CD = \frac{2 - 3}{-3 - 3} = \frac{-1}{-6} = \frac{1}{6}$.
Since the slope of $AB = \text{slope of } CD$,$AB$ is parallel to $CD$.
Now,the slope of $BC = \frac{3 - 0}{3 - 4} = \frac{3}{-1} = -3$.
The slope of $AD = \frac{2 - (-1)}{-3 - (-2)} = \frac{3}{-1} = -3$.
Since the slope of $BC = \text{slope of } AD$,$BC$ is parallel to $AD$.
Since both pairs of opposite sides are parallel,the quadrilateral $ABCD$ is a parallelogram.
The slope of $AB = \frac{0 - (-1)}{4 - (-2)} = \frac{1}{6}$.
The slope of $CD = \frac{2 - 3}{-3 - 3} = \frac{-1}{-6} = \frac{1}{6}$.
Since the slope of $AB = \text{slope of } CD$,$AB$ is parallel to $CD$.
Now,the slope of $BC = \frac{3 - 0}{3 - 4} = \frac{3}{-1} = -3$.
The slope of $AD = \frac{2 - (-1)}{-3 - (-2)} = \frac{3}{-1} = -3$.
Since the slope of $BC = \text{slope of } AD$,$BC$ is parallel to $AD$.
Since both pairs of opposite sides are parallel,the quadrilateral $ABCD$ is a parallelogram.
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Similar Questions
In the triangle with vertices at $A(6,3), B(-6,3)$ and $C(-6,-3)$,the median through $A$ meets $BC$ at $P$,the line $AC$ meets the $x$-axis at $Q$,while $R$ and $S$ respectively denote the orthocentre and centroid of the triangle. Then the correct matching of the coordinates of points in List-$I$ to List-$II$ is:
$i$. $P$ $A$. $(0,0)$ $ii$. $Q$ $B$. $(6,0)$ $iii$. $R$ $C$. $(-2,1)$ $iv$. $S$ $D$. $(-6,0)$ $E$. $(-6,-3)$ $F$. $(-6,3)$
| $i$. $P$ | $A$. $(0,0)$ |
| $ii$. $Q$ | $B$. $(6,0)$ |
| $iii$. $R$ | $C$. $(-2,1)$ |
| $iv$. $S$ | $D$. $(-6,0)$ |
| $E$. $(-6,-3)$ | |
| $F$. $(-6,3)$ |
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