Two sides of a triangle have joint equation : x²-2xy-3y²+8y-4=0, the third side which is variable always passes through the point (-5,-1). Find range of values of slope of third side such that origin is an interior point.

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Bhavyaa · Answer

(-1,0.2) range Two given lines are x+y=2 and x=3y-2 Draw a rough idea y=me+5m-1 be eq of third side They form a triangle if slope is greater than -1 and the third side will contain origin for m=0.2

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Two sides of a triangle have joint equation : x²-2xy-3y²+8y-4=0, the third side which is variable always passes through the point (-5,-1). Find range of values of slope of third side such that origin is an interior point.

Two sides of a triangle have joint equation : x²-2xy-3y²+8y-4=0, the third side which is variable always passes through the point (-5,-1). Find range of values of slope of third side such that origin is an interior point.

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Two sides of a triangle have joint equation : x²-2xy-3y²+8y-4=0, the third side which is variable always passes through the point (-5,-1). Find range of values of slope of third side such that origin is an interior point.

Two sides of a triangle have joint equation : x²-2xy-3y²+8y-4=0, the third side which is variable always passes through the point (-5,-1). Find range of values of slope of third side such that origin is an interior point.

1 Answers

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