Question

a quadrilateral has
vertices of a parallelogram.
21 A quadrilateral has vertices (4, 1), (1,7), (-60) and (-1, - 9). Show that the mid-points of
the sides of this quadrilateral form a parallelogram,​

Answer 1

For More ANSWER Follow me Dude's ⏬....Now..

MARK IN BRAIN LIST

✨✨

Answer 2

$\huge\pink{\boxed{\blue{\boxed{ \purple{ \boxed{{\pink{Answer}}}}}}}} \\ \large\pink{\boxed{\blue{\boxed{ \purple{ \boxed{{\pink{Your~answer↓}}}}}}}}$

$\boxed{ \large{ \mathfrak{Given :- }}}$

A quadrilateral having vertices (4, 1), (1,7), (-60) and (-1, - 9).

$\boxed{ \large{ \mathfrak{To Show :- }}}$

The mid-points ofthe sides of this quadrilateral form a parallelogram.

$\large\bold\red{Formula \: used } \\ \boxed{ \large{ {Slope \: of \: line \: = \frac{y_2 - y_1}{x_2 - x_1 } }}}$

Solution :-

Let us suppose a quadrilateral ABCD having vertices A(4, 1), B(1, 7), C(- 6, 0) and D(- 1, -9)

Let P, Q, R and S be the midpoints of AB, BC, CD & DA respectively.

$Coordinates \: of \: P \\ = Midpoint \: of \: AB \\ = ( \frac{4 + 1}{2} , \frac{1 + 7}{2} ) \\ = ( \frac{5}{2} , 4)$

$Coordinates \: of \: Q \\ = Midpoint \: of \: BC \\ = ( \frac{ - 6 + 1}{2} , \frac{0 + 7}{2} ) \\ = ( \frac{ - 5}{2} , \frac{7}{2} )$

$Coordinates \: of \: R \\ = Midpoint \: of \: CD \\ = ( \frac{ - 6 - 1}{2} , \frac{0 - 9}{2} ) \\ = ( \frac{ - 7}{2} , \frac{ - 9}{2} )$

$Coordinates \: of \: S\\ = Midpoint \: of \: AD \\ = ( \frac{4 - 1}{2} , \frac{1 - 9}{2} ) \\ = ( \frac{3}{2} , - 4)$

$Slope of PQ = \frac{y_2 - y_1}{x_2 - x_1 } \\ = \frac{ \frac{7}{2} - 4 }{ \frac{ - 5}{2} - \frac{5}{2} } \\ = \frac{ \frac{ - 1}{2} }{ \frac{ - 10}{2} } = \frac{1}{10}$

$Slope of RS = \frac{y_2 - y_1}{x_2 - x_1 } \\ = \frac{ - 4 - \frac{ - 9}{2} }{ \frac{3}{2} - \frac{ - 7}{2} } \\ = \frac{ \frac{1}{2} }{ \frac{10}{2} } = \frac{1}{10}$

Since, slope of PQ = Slope of RS

∴ PQ || RS

$Slope of QR= \frac{y_2 - y_1}{x_2 - x_1 } \\ = \frac{ \frac{7}{2} - \frac{ - 9}{2} }{ \frac{ - 5}{2} - \frac{ - 7}{2} } \\ = \frac{ \frac{16}{2} }{ \frac{2}{2} } = 8$

$Slope of PS= \frac{y_2 - y_1}{x_2 - x_1 } \\ = \frac{4 - ( - 4)}{ \frac{5}{2} - \frac{3}{2} } \\ = 8$

Since, Slope of QR = Slope of PS

∴ QR || PS

Now, PQ || RS & QR || PS

∴ PQRS is a parallelogram.

$\boxed{ \huge{ \mathfrak{Mark \: me \: as \: Brainliest}}}$