Question

Show that point P(2,-2),Q(7,3),R(11,-1) and S(6,-6) are vertices of a parallelogram

Answer 1

Here is the answer

Solution
P(2 ,-2)
Q(7 ,3)
R(11 ,-1)
S(6 ,-6)

Let's find out the slopes of the lines

Slope of line PQ

$\bf{= \frac{3 - ( - 2)}{7 - 2} = \frac{3 + 2}{5} = \frac{5}{5} = 1}$

✔️This is statement (1)



Slope of line QR

$\bf{= \frac{ - 1 - 3}{11 - 7} = \frac{ - 4}{4} = - 1}$

✔️Statement (2)



Slope of line RS


$\bf{ = \frac{ - 6 - (- 1)}{6 - 11} = \frac{ - 6 + 1}{ - 5} = \frac{ - 5}{ - 5} = 1}$

✔️Statement (3)



Slope of line PS

$\bf{=\frac{ - 6 - ( -2 )}{6 - 2} = \frac{ - 6 + 2}{4} = \frac{ - 4}{4} = - 1}$

✔️Statement (4)

Thus,
From (1) and (3),
Slope of line PQ = Slope of line RS

Therefore,
Line PQ || Line RS

Remember that,

If two lines have equal slopes , then they are parallel.

Also,

From (2) and (4) ,
Slope of line QR = Slope of line PS
°•° Line QR || Line PS.
Also,

If two lines have equal slopes , then they are parallel.

Therefore,

$\therefore \square{ABCD} \: is \: a \: Parallelogram$

Hence,
Point P(2,-2),Q(7,3),R(11,-1) and S(6,-6) are vertices of a parallelogram

Answer 2

mid point of diagonal PR={(2+11)/2,(-2-2)/}=(13/2,-3/2)
[ from midpoint section formula , we know that if two points( x1, y1) and ( x2,y2) are given then midpoint of line joining of give points 15 { (x1+ x2)/2,( y1+ y2)/2}]
similarly,midpoint of QS ={(7+6)/2,(3-6)/2}=(13/2) here we see midpoint of PR= midpoint of QS
so,PQRS is parralellogram