Skip to main content
Back to STRAIGHT LINES

JEE MATHEMATICS MCQ: STRAIGHT LINES

1 Easy
1 of 1
Application +4 | -1
If the lines x=a+m,y=−2x=a+m, y=-2 and y=mxy=m x are concurrent, the least value of ∣a∣|a| is :
View answer and explanation
Correct option: C
222 \sqrt{2}
Option C (222 \sqrt{2}) is correct because the condition for concurrency leads to a quadratic equation in 'm' which must have real roots, thus giving the least value of |a| as 222 \sqrt{2}.

More Questions from STRAIGHT LINES

Practice All Chapter MCQs →
Type at least 2 characters or jump directly:
🔓

Save your learning progress

Sign in with a free account to save your practice answers and quiz results, then return to them when you revise.