Question

P(3,1),Q(6,5) and R(x,y) are three points such that the angle PRQ is a right angle and the area of △RQP=7, then 4x-3y+5 is equals to
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Answer 1

Answer:

☞Given :-

ar(ΔRQP)=7

P=(3,1)

Q=(6,5) And,R=(x,y)

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Now,ar(ΔRQP)=

21 ∣x 1 (y 2 −y 3 )+x 2 (y 3 −y 1 )+x 3 (y 1 −y 2 )∣

⇒7= 21

∣3(5−y)+6(y−1)+x(1−5)∣ [On putting the value of points]

⇒7×2=15−3y+6y−6+x−5x

⇒14=9+3y−4x

⇒4x−3y=9−14

⇒4x−3y=−5

∴4x−3y+5=0

• I hope it was helpful to you :)

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Answer 2

Answer:

ar(ΔRQP)=7

P=(3,1)

Q=(6,5) And,R=(x,y)

Now,ar(ΔRQP)=

21 ∣x 1 (y 2 −y 3 )+x 2 (y 3 −y 1 )+x 3 (y 1 −y 2 )∣

⇒7= 21

∣3(5−y)+6(y−1)+x(1−5)∣ [On putting the value of points]

⇒7×2=15−3y+6y−6+x−5x

⇒14=9+3y−4x

⇒4x−3y=9−14

⇒4x−3y=−5

∴4x−3y+5=0