JEE MATHEMATICS MCQ: STRAIGHT LINES

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Analyzing +4 | -1

For

a>b>c>0,

the distance between

(1,1)

and the point of intersection of the lines

a x+b y+c=0

and

b x+a y+c=0

is less than

2 \sqrt{2}.

Then

View answer and explanation

Correct option: A

a+b-c>0

Option A (

a+b-c>0

) is correct because it is derived from the distance formula and the given condition that the distance between (1,1) and the intersection point is less than

2\sqrt{2}

More Questions from STRAIGHT LINES

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If the three lines

x-3 y=p, a x+2 y=q

and

a x+y=r

form a right-angled triangle then :
The orthocentre of the triangle whose vertices are

\left\{ {a{t_1}{t_2},\;a\left( {{t_1} + {t_2})\} ,\{ a{t_2}{t_3},a({t_2} + {t_3}} \right)} \right\},\;\left\{ {a{t_3}{t_1},\;a\left( {{t_3} + {t_1}} \right)} \right\}

is
The distance between the parallel lines

y = 2\;x + 4

and

6x = 3y + 5

is The distance between the pair of lines given by

{x^2} + {y^2} + 2xy - 8ax - 8ay - 9{a^2} = 0

is Find the image of the point

(3,8)

with respect of the line

x+3 y=7

assuming the line to be plane mirror: The centroid of the triangle whose three sides are given by the combined equation

\left( {{x^2} + 7xy + 2{y^2}} \right)\left( {y - 1} \right) = 0