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प्रश्न
Find the type of the quadrilateral if points A(–4, –2), B(–3, –7) C(3, –2) and D(2, 3) are joined serially.
उत्तर
The given points are A(–4, –2), B(–3, –7) C(3, –2) and D(2, 3).
If they are joined serially so,
\[\text{Slope of AB} = \frac{y_{2}-y_{1}}{x_{2}-x_{1}}\]
= \[\frac{-7 - (-2)}{-3 - (-4)}\]
= \[\frac{- 7 + 2}{- 3 + 4} = - 5\]
\[\text{Slope of BC} = \frac{y_2-y_1}{x_2-x_1}\]
= \[\frac{-2-(-7)}{3-(-3)}\]
= \[\frac{- 2 + 7}{3 + 3} = \frac{5}{6}\]
\[\text{Slope of CD} = \frac{y_2-y_1}{x_2-x_1}\]
= \[\frac{3-(-2)}{2-3}\]
= \[\frac{3 + 2}{2 - 3} = - 5\]
\[\text{Slope of AD} = \frac{y_2-y_1}{x_2-x_1}\]
= \[\frac{3-(-2)}{2-(-4)}\]
= \[\frac{3 + 2}{2 + 4} = \frac{5}{6}\]
Slope of AB = slope of CD
∴ line AB || line CD
Slope of BC = slope of AD
∴ line BC || line AD
Both the pairs of opposite sides of ∆ABCD are parallel.
∴ ABCD is a parallelogram.
∴ The quadrilateral formed by joining the points A, B, C and D is a parallelogram.
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Solution:
Slope of line = `("y"_2 - "y"_1)/("x"_2 - "x"_1)`
∴ Slope of line AB = `(2 - 1)/(8 - 6) = square` .......(i)
∴ Slope of line BC = `(4 - 2)/(9 - 8) = square` .....(ii)
∴ Slope of line CD = `(3 - 4)/(7 - 9) = square` .....(iii)
∴ Slope of line DA = `(3 - 1)/(7 - 6) = square` .....(iv)
∴ Slope of line AB = `square` ......[From (i) and (iii)]
∴ line AB || line CD
∴ Slope of line BC = `square` ......[From (ii) and (iv)]
∴ line BC || line DA
Both the pairs of opposite sides of the quadrilateral are parallel.
∴ `square`ABCD is a parallelogram.
Show that points A(– 4, –7), B(–1, 2), C(8, 5) and D(5, – 4) are the vertices of a parallelogram ABCD
