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Question

The combined equation of two sides of a triangle is x23y22xy+8y4=0. The third side, which is variable always passes through the point (5,1). If the range of values of the slope of the third line such that the origin is an interior point of the triangle is (a,b), then the value of (a+1b) is

A
12
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B
2
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C
0
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D
4
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Solution

The correct option is D 4
x23y22xy+8y4=0(x3y)(x+y)+2x+2y2x+6y4=0(x+y)(x3y+2)2(x3y+2)=0(x+y2)(x3y+2)=0
So, the lines are x+y2=0 and x3y+2=0
Let the equation of the variable line be
y+1=m(x+5)


For origin to lie inside the required triangle y intercept for the variable line should be ive
5m1<0m<15
(5,1) also lies on x3y+2=0
Now triangle can form when
m>1
So, m(1,1/5)
a+1b=4

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