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points A and B are in the first quadrant. point Ois the origin.If the slope of OA is 1, slope of OB is 7 and OA= OB, then slope of AB is -------
Hi Deepu,
Draw the figure.
Let A be angle of OA with x-axis, and B be the angle of OB with x-axis.
Let m be the required slope. Let Q be the (in acute sense) angle made by AB with horizontal.
So m = -tanQ
Now tanA=1, or SinA=1/root2 and cosA=1/root2
and tanB=7, or SinB=7/root50 and cosB=1/root50.
From figure, tanQ = (sinB-sinA)/(cosA-cosB)
or tanQ = (7root2 - root50)/(root50 - root2) = (7-5)/(5-1) = 1/2
or required slope m = -1/2.
Regards,
Ashwin (IIT Madras).
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