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.
Then

Options

A. angle at A is acute
B. angle at C is acute
C. internal angle bisector at angle B is 3x+6y-16=0
D. external angle bisector at angle C is 8x+8y+7 = 0

Answer

A, C, D

Explanation

The slopes of the line AB, BC and CA are -1, `-(1)/(7)` and -7, respectively,
`"Let " m_(1) = -(1)/(7), m_(2) = -1, m_(3) =-7.`
`therefore m_(1) gt m_(2) gt m_(3)`
So, tangent fo internal angles of the triangle are
`"tan" A = (3)/(4), "tan" B = (3)/(4) " and tan" C = -(24)/(7)`
So, interior angles A and B are acute and interior angle C is abtuse.
`therefore " Internal bisector of B "-= " Acute angle bisector at B"`

`" " -= " Acute angle bisector of lines AB and BC "`
`" " -= 3x+6y-16=0`
External bisector of C `-= " Acute bisector of C" `
` " " -= " Acute angle bisector of lines AC and BC"`
` " " -=8x+8y+7=0`