(7 p q r - 5 ²) - (Pqr +5)2 = -35
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(pqr +1)2 - (pqr-1)2
Answers
Step-by-step explanation:
In triangle PQR, A (-2, 3), B (5, -1) and C (-4, -7) are the mod points of PQ, QR and PR respectively. What are the coordinates of P, Q and R?
In triangle PQR, A (-2, 3), B (5, -1) and C (-4, -7) are the mid points of PQ, QR and PR respectively. What are the coordinates of P, Q and R?
P = A+C-B = (−11,−3)
Q = A+B-C = (7,9)
R = B+C-A = (3,−11)
Let P(x₁,y₁) , Q(x₂, y₂) and R(x₃, y₃) be the vertices of the triangle
Since A(-2, 3), B(5, -1) and C(-4, -7) are the mid points of PQ QR and PR
Hence
(x₁+ x₂)/2=-2 i. e. x₁+ x₂=-4
(x₂+x₃)/2= 5 i.e. x₂+x₃=10
(x₁ +x₃ )/2=-4 i.e. x ₁+x₃=-8
Adding we get
2(x₁+ x₂+x₃)=-2. Or x₁+ x₂+x₃=-1
Hence x₃=3, x₁=-11, x₂=7
Similarly (y₁+y₂)/2=3 I. e y₁+y₂=6
( y₂+ y₃) /2=-1 I.e. y₂+ y₃=-2
and. (y₁+ y₃) /2=- 7. I.e. y₁+ y₃=-14
Solving gives y₁=6 , y₂=6 and y₃=-11
Hence P(-11, -3) , Q(7, 9) and R(3, -11) are the vertices of the triangle