NTA Abhyas JEE Main2020MathematicsStraight LinesPractice
Let the incentre of Δ A B C is Ι 2,5 . If A = 1,13 and B = - 4,1 , then the coordinates of C are
Options
- A1,10
- B10,1
- C8,2
- D9,3
Correct answer
B. 10,1
Step-by-step solution
Slope of A B is 13 - 1 1 - - 4 = 12 5 Slope of B I is 5 - 1 2 - - 4 = 2 3 Slope of A I is 13 - 5 1 - 2 = - 8 Let, the slopes of B C and A C are m 1 & m 2 respectively ⇒ m 2 - - 8 1 + m 2 - 8 = - 8 - 12 5 1 + - 8 12 5 m 2 + 8 91 = 1 - 8 m 2 - 57 8 × 91 - 57 = 8 × 57 - 91 m 2 ⇒ m 2 = - 4 3 Now, m 1 - 2 3 1 + m 1 2 3 = 2 3 - 12 5 1 + 2 3 × 12 5 ⇒ 3 m 1 - 2 39 = 3 + 2 m 1 - 26 ⇒ 9 m 1 - 6 = - 6 - 4 m 1 ⇒ m 1 = 0 ⇒ The equation of A C is y - 13 = - 4 3 x - 1 and the equation of B C is y = 1 ⇒ C is 10,1