Question

24. Find the incentre of the triangle formed by the straight lines x + 1 = 0, 3x - 4y = 5,
5x + 12y = 27.​

Answer 1

Answer:

⟶ Equation of sides ⇒

12x

2

−20xy+7y

2

=0

12x

2

−6xy−14xy+7y

2

=0

6x(2x−y)−7y(2x−y)=0

(6x−7y)(2x−y)=0

for B→ we need to solve

6x−7y=0 & 2x−3y+4=0

y=

7

6x

substituting y in 2x−3y+4=0

2x−3(

7

6x

)+4=0

2x−

7

18x

+4=0⇒14x−18x+28=0

4x=28

x=7

y=

7

6x

=

7

6

7=6

∴B=(7,6)

for C→ we need to solve

2x−3y+4=0 & 2x−y=0

y=2x

substituting y=2x in 2x−3y+4=0

2x−3(2x)+4=0

2x−6x+4=0

4x=4

x=1

y=2x

=2(1)

y=2

∴C≡(1,2)