a quadratic eq whose roots are the x,y intercepts of the line passing through (1,1) and making a triangle of area A with the axes may be?
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Answer:
x
2
−2Ax+2A=0
Line passing through (1,1) is
y−1=m(x−1)
y=mx+1−m
x intercept : x=
m
m−1
---(1)
y=0
y intercept : y=1−m ---(2)
x=0
Given area of OAB=A=
2
1
×(OB)(OA)
=
2
1
×(1−m)
m
(m−1)
A=
2m
−(m
2
−1)
---(3)
Quadratic equation whose roots are x,y is
x
2
−(x+y)x
1
+xy=0
x
1
2
−(
m
2m−m
2
−1
)x
1
−
m
(m−1)
2
=0
x
1
2
+(
m
m
2
−2m+1
)x
1
−
m
(m−1)
2
=0
From 3, we get
x
1
2
−2Ax
1
+2A=0
Step-by-step explanation:
Hope it helps
Brainlist Mark krdo plz
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