JEE MATHEMATICS MCQ: STRAIGHT LINES

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Application +4 | -1

The incentre of the triangle formed by (0, 0), (5, 12), (16, 12) is

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Correct option: A

(7, 9)

Option A (7, 9) is correct because the incenter is calculated using the formula

\left( {\frac{{a{x_1} + b{x_2} + c{x_3}}}{{a + b + c}},\;\frac{{a{y_1} + b{y_2} + c{y_3}}}{{a + b + c}}} \right)

, where

a, b, c

are the side lengths of the triangle.

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