Point Equidistant from Three Lines (Jee Mains) | ExamDuo
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Triangle Geometry and Bisectors
Hard
Point Equidistant from Three Lines
A point equidistant from the lines x + √
3
y + 4 = 0, √
13
x + 6y + 14 = 0 and 7x + 24y – 50 = 0 is
SELECT ONE OPTION
A
(1 , -1)
B
(1 , 1)
C
(0 , 0)
D
(0 , 1)
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