in a ∆abc, if a is the point (1, 2) and equations of medians through b and c are respectively x+y=5 and x=4 then points of b are
Answers
ABC is a triangle in which A(1,2). Mid point of AC is D and mid point of AB is E.
Equation of mdn BD is x+y=4… ….(1).
Equation of mdn CE is x=4…………(2)
let coordinate of B(p,q) and C(r,s)
coordinate of E [(p+1)/2,(q+2)/2]
Point E lies on CE [x=4 ] , therefore
(p+1)/2=4
p+1=8
p=8–1
p=7
Point B(p,q) lies on BD [x+y=4].
p+q=4
7+q=4
q=-3 . Theredore B(7,-3) , Answer.
Point C(r,s) lies on CE[ x = 4]
r=4
coordibate of D[(r+1)/2,(s+2)/2]
Similarly point D lies on BD[x+y=4].
(r+1)/2+(s+2)/2=4
r+s=5 , put r = 4
4+s=5 =>s=1 , therefore C(4,1) , Answer.
Answer:
ΔABC has vertex A(1,2)
Given that midpoint of AC is D and mid point of AB is E
Also,
equation of medium BD is x+y=5
equation of medium CE is x=4
let coordinate of B(p,q) and C(r,s)
∴ coordinate of E≡(
2
p+1
,
2
q+1
)
But E lies on the line segment CE (x=4)
∴
2
P+1
=4
⇒P=7
And point B(p,q) lies on the line BD [x+y=4]
∴ p+q=4
⇒q=−3
∴ B≡(7,−3)
Point C(r,s) lies on CE [x=4]
∴ r=4
Coordinates of D are ≡[
2
r+1
,
2
s+2
]
Similarly point D lies on BD [x+y=5]
∴
2
r+1
+
2
s+2
=5
r+s=7
⇒s=3
∴ C≡(4,3)
∴ coordinates of B & C are (7,−3) & (4,3)
Step-by-step explanation:
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