A straight line L with negative slope passes through the point (8,2) and cuts positive coordinate axes at the points P and Q. Find the minimum value of OP + OQ as L varies, where O is the origin
class :- 11th
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Since the slope is negative, let the slope of line be −m where m is positive.
The equation of line is y−2=−m(x−8)
y+mx=2+8m
Now the y intercept is 2+8m
and the x intercept is 8+
m
2
Sum is equal to 10+8m+
m
2
The minimum value of 8m+
m
2
is 8. (By AM,GM inequality).
The minimum value of intercepts is 18
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