JEE MATHEMATICS MCQ: STRAIGHT LINES

1 Hard

1 of 1

Application +4 | -1

The incentre of a triangle with vertices (7, 1),

\left( { - 1,\;5} \right)

and

\left( {3 + 2\sqrt 3 ,\;3 + 4\sqrt 3 } \right)

View answer and explanation

Correct option: A

\left( {3 + \frac{2}{{\sqrt 3 }},3 + \frac{4}{{\sqrt 3 }}} \right)

Option A,

\left( {3 + \frac{2}{{\sqrt 3 }},3 + \frac{4}{{\sqrt 3 }}} \right)

, is correct because the triangle is equilateral, and the incenter coincides with the centroid, which can be calculated by averaging the coordinates.

More Questions from STRAIGHT LINES

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

a x+y=r

\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\}

y = 2\;x + 4

6x = 3y + 5

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

(3,8)

x+3 y=7

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