Question

The straight lines $x+y-4=0,3x+y+7=0$ and $x+3y-10=0$ forms a triangle which is

• A. isosceles
• B. right angled
• C. equilateral
• D. obtuse triangle

Answer 1

Answer: (A), (D)

The correct options are
A isosceles
D obtuse triangle

One method to discuss the nature of $△ABC$ is to determine the points of intersection of lines which gives the co-ordinates of vertices $A,B$ and $C$ and then find the length $AB,BC$ and $AC$.

Another method is to find the internal angles of $△ABC$.

Slopes of sides $AB,BC$ and $CA$ are $-\frac{1}{3},-1,-3$

respectively which is indicated in figure .

Therefore $\angle A={\tan }^{-1}\left(\frac{-3+1/3}{1+1}\right)={\tan }^{-1}(-\frac{4}{9})[$ obtuse angle $]$

$\angle B={\tan }^{-1}\left(\frac{-1/3+1}{1+1/3}\right)={\tan }^{-1}\left(\frac{1}{2}\right),;$

$\angle C={\tan }^{-1}\left(\frac{-1+3}{1+3}\right)={\tan }^{-1}\left(\frac{1}{2}\right)$

$△ABC$ is both isosceles and obtuse.