Question

Challenge if you will Solve I will follow u, thank u and Mark as brainlist... super hard question

Answer 1

express the equation x/3+y/6=1 in the form of ax+by+c=0 and indicate the values of a,b,c

Answer 2

$Here \: is \: the \: answer \: of \: your \: question$

$\underline{Given}$ :- P, Q, R and S ae the mid points of side AB, BC, CD and DA pf rhombus ABCD.

$\underline{To}$ $\underline{Prove}$ :- ABCD is a rectangle.

$\underline{Construction}$ :- Join BD.

$\underline{Proof}$ :-

In ∆ ABD

P is the mid point of AB.

and S is the mid point of AD.

So, PS ll BD .......(1)

And PS = ½ BD ........(2)

Also in ∆ BDC.

R and Q are the mid points of sides CD and BC.

So, RQ ll BD .......(3)

And RQ = ½ BD ........(4)

From (1) and (3)

PS ll RQ

And from (2) and (4)

PS ll RQ

In PQRS

PS ll RQ and PS = RQ.

So, PQRS is a parallelogram.

* What we have to prove is that PQRS is a rectangle.

Now,

DC = BC [As sides of rhombus are equal]

Then, their half's are also equal.

½ DC = ½ BC

RC = CQ [As they are mid points of DC and BC]

Now, In ∆ RQC.

RC = CQ

Angle 1 = Angle 2 [Opposite sides of parallelogram are equal] .......(5)

In ∆SDR and ∆ BQP

DR = BQ [DC = BC; ½ DC = ½ BC]

DS = BP [AD = AB; ½ AD = ½ AB]

SR = PQ [Opposite sides of parallelogram are equal]

∆ SDR ~ ∆ BQP

Angle 3 = Angle 4 ......(6)

By Cpct.

Now, DC is a line.

Then,

Angle 3 + Angle SRQ + Angle 1 = 180° .......(7)

Similarly, BC is also a line.

Angle 2 + Angle RQP + Angle 4 = 180°

Angle 1 + Angle RQP + Angle 3 = 180° .....(8)
[From (5) and (6)]

Now, add (7) and (8)

Angle 1 + Angle RQP + Angle 3 = Angle 3 + Angle SRQ + Angle 1

Angle RQP = Angle SRQ .......(9)

Now,

SR ll PQ and RQ is a transversal.

Angle SRQ + Angle RQP = 180°

Angle SRQ + Angle SRQ = 180° [From (9)]

2 (Angle SRQ) = 180°

Angle SRQ = 90°

So, PQRS is a parallelogram with angle 90°.

So,

$\textbf{PQRS is a rectangle}$
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