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Question
Equation of straight line ax+by+c=0, where 3a+4b+c=0, which is at maximum distance from (1,−2),is
Equation of straight line ax+by+c=0, where 3a+4b+c=0, which is at maximum distance from (1,−2),is
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Solution
The correct option is C x+3y−15=0
It passes through a fixed point (3,4)
slope of line joining (3,4) and (1,−2) is −6−2=3For the straight line to be at a maximum distance from the point (1,−2), it should be perpendicular to the line joining (3,4) and (1,−2).
∴ slope of required line = −13
equation is y−4 = −13(x−3)
x+3y−15=0
It passes through a fixed point (3,4)
slope of line joining (3,4) and (1,−2) is −6−2=3
For the straight line to be at a maximum distance from the point (1,−2), it should be perpendicular to the line joining (3,4) and (1,−2).
∴ slope of required line = −13
equation is y−4 = −13(x−3)
x+3y−15=0
∴ slope of required line = −13
equation is y−4 = −13(x−3)
x+3y−15=0
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