A triangle is formed by the lines whose equations are AB : x + y – 5 = 0, BC :
x + 7y – 7 = 0 and CA : 7x + y + 14 = 0. Find the bisector of the interior angle at B and the exterior angle at C. Determine the nature of the interior angle at A and find the equation of the bisector.
Text Solution
Verified by ExpertsCHECK THE SOLUTION.
3x + 6y – 16 = 0; 8x + 8y + 7 = 0 ; 12x + 6y – 11 = 0
Sol. The slopes of the lines AB, BC and CA are –1, – 
and –7 respectively
Let m 1 = – 
, m 2 = –1, m 3 = – 7 m 1 > m 2 > m 3
∴ tangent of internal angles of the triangle are
tanA = 
, tanB = 
, tanC = 
⇒ tan A = 
, tan B = 
and tan C = – 
∴ interior angles A and B are acute and interior angle C is obtuse
∴ internal bisector of B = acute bisector of B ≡ 3x + 6y – 16 = 0
External bisector of C ≡ acute bisector of C ≡ 8x + 8y + 7 = 0
Internal bisector of A ≡ acute bisector of A ≡ 12x + 6y – 11 = 0
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