Question

A triangle is formed by the lines whose equations are $A B: x+y-5=0, B C: x+7 y-7=0$ and $C A: 7 x+y+14=0$. Then

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

Answer 1

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

The slopes of the lines $A B, B C$ and $C A$ 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
$\tan A=\frac{3}{4}, \tan B
=\frac{3}{4} \text { and } \tan C
=-\frac{24}{7}$
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 $A B$ and $B C$
$\equiv 3 x+6 y-16=0$
External bisector of $C \equiv$ Acute bisector of $C$
$\equiv$ Acute angle bisector of lines $A C$ and $B C$ $\equiv 8 x+8 y+7=0$