Question

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

• 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 1

Answer: (A), (C), (D)

The slopes of the lines $AB,BC$ and $CA$ are $-1,-\frac{1}{7}$ and $-7$, respectively
Let $m_1=-\frac{1}{7},m_2=-1,m_3=-7$.
$\therefore m_1>m_2>m_3$
So, tangent of internal angles of the triangle are
$\tan A=\frac{3}{4},\tan B<br/>=\frac{3}{4}\text{ and }\tan C<br/>=-\frac{24}{7}$
So, interior angles $A$ and $B$ are acute and interior angle $C$ is obtuse.
$\therefore$ Internal bisector of $B\equiv$ Acute angle bisector at $B$
$\equiv$ Acute angle bisector of lines $AB$ and $BC$
$\equiv 3x+6y-16=0$
External bisector of $C\equiv$ Acute bisector of $C$
$\equiv$ Acute angle bisector of lines $AC$ and $BC$ $\equiv 8x+8y+7=0$