Question

The straight lines $x + y = 0$, $3x + y - 4 = 0$, $x + 3y - 4 = 0$ form a triangle which is
1) Isosceles
2) Equilateral
3) Right Angled
4) None of these

Answer 1

Hint: From the equations of the given lines of the triangle we can find the slopes of the triangle. Taking the slopes of two given lines at a time we can find the tangent of the angle between the two lines of the triangle by the formula,
\[\tan \theta = \left| {\dfrac{{{m_1} - {m_2}}}{{1 + {m_1}{m_2}}}} \right|\]
Where, ${m_1},{m_2}$ is the slope of two lines respectively
From the angles of the triangles, we can get a clear idea about the sides of the triangle, because we know, if two angles of a triangle are equal, then the two sides opposite to the two corresponding angles are equal to.

Complete step-by-step answer:
Given lines of the triangle are,
$x + y = 0 - - - \left( 1 \right)$
$3x + y - 4 = 0 - - - \left( 2 \right)$
$x + 3y - 4 = 0 - - - \left( 3 \right)$
The general equation of a line is, $y = mx + c$.
Where, $m = $slope of the line.
We can write $\left( 1 \right)$ as,
$x + y = 0$
$ \Rightarrow y = - x$
Now, comparing it to the general equation of a line, we get,
Slope of $\left( 1 \right)$, ${m_1} = - 1$
We can write $\left( 2 \right)$ as,
$3x + y - 4 = 0$
$ \Rightarrow y = - 3x + 4$
Now, comparing it to the general equation of a line, we get,
Slope of $\left( 2 \right)$, ${m_2} = - 3$
We can write $\left( 3 \right)$ as,
$x + 3y - 4 = 0$
$ \Rightarrow y = - \dfrac{1}{3}x + \dfrac{4}{3}$
Now, comparing it to the general equation of a line, we get,
Slope of $\left( 3 \right)$, ${m_3} = - \dfrac{1}{3}$
Now, we know, the tangent of the angle between two lines is given by,
\[\tan \theta = \left| {\dfrac{{{m_1} - {m_2}}}{{1 + {m_1}{m_2}}}} \right|\]
Let the angle between lines $\left( 1 \right)$ and $\left( 2 \right)$, $\left( 2 \right)$ and $\left( 3 \right)$, $\left( 3 \right)$ and $\left( 1 \right)$ is ${\theta _1},{\theta _2},{\theta _3}$ respectively.
Therefore, the tangent of the angle between lines $\left( 1 \right)$ and $\left( 2 \right)$ is,
$\tan {\theta _1} = \left| {\dfrac{{\left( { - 1} \right) - \left( { - 3} \right)}}{{1 + \left( { - 1} \right)\left( { - 3} \right)}}} \right|$
$ \Rightarrow \tan {\theta _1} = \left| {\dfrac{{ - 1 + 3}}{{1 + 3}}} \right|$
Now, simplifying, we get,
$ \Rightarrow \tan {\theta _1} = \left| {\dfrac{2}{4}} \right|$
$ \Rightarrow \tan {\theta _1} = \left| {\dfrac{1}{2}} \right| = \dfrac{1}{2}$
The tangent of the angle between lines $\left( 2 \right)$ and $\left( 3 \right)$ is,
$\tan {\theta _2} = \left| {\dfrac{{\left( { - 3} \right) - \left( { - \dfrac{1}{3}} \right)}}{{1 + \left( { - 3} \right)\left( { - \dfrac{1}{3}} \right)}}} \right|$
$ \Rightarrow \tan {\theta _2} = \left| {\dfrac{{ - 3 + \dfrac{1}{3}}}{{1 + 1}}} \right|$
Now, simplifying, we get,
$ \Rightarrow \tan {\theta _2} = \left| {\dfrac{{\dfrac{{ - 9 + 1}}{3}}}{2}} \right|$
$ \Rightarrow \tan {\theta _2} = \left| {\dfrac{{ - 8}}{{3 \times 2}}} \right|$
$ \Rightarrow \tan {\theta _2} = \left| {\dfrac{{ - 4}}{3}} \right| = \dfrac{4}{3}$
[We take the positive value because, we are using the modulus function]
Now, the tangent of the angle between lines $\left( 1 \right)$ and $\left( 3 \right)$ is,
$\tan {\theta _3} = \left| {\dfrac{{\left( { - 1} \right) - \left( { - \dfrac{1}{3}} \right)}}{{1 + \left( { - 1} \right)\left( { - \dfrac{1}{3}} \right)}}} \right|$
$ \Rightarrow \tan {\theta _3} = \left| {\dfrac{{ - 1 + \dfrac{1}{3}}}{{1 + \dfrac{1}{3}}}} \right|$
Now, simplifying, we get,
$ \Rightarrow \tan {\theta _3} = \left| {\dfrac{{\dfrac{{ - 3 + 1}}{3}}}{{\dfrac{{3 + 1}}{3}}}} \right|$
$ \Rightarrow \tan {\theta _3} = \left| {\dfrac{{\dfrac{{ - 2}}{3}}}{{\dfrac{4}{3}}}} \right|$
Multiplying the numerator and denominator by $\dfrac{3}{2}$, we get,
$ \Rightarrow \tan {\theta _3} = \left| {\dfrac{{ - \dfrac{1}{1}}}{{\dfrac{2}{1}}}} \right| = \left| {\dfrac{{ - 1}}{2}} \right| = \dfrac{1}{2}$
[We take the positive value because, we are using the modulus function]
Therefore, we got, $\tan {\theta _1} = \tan {\theta _3}$
$ \Rightarrow {\theta _1} = {\theta _3}$
We know, if two angles of a triangle are equal, then the two sides opposite to the two corresponding angles are equal to.
Therefore, since two angles are equal, two sides of the triangle are also equal to each other.
Hence, the triangle is an isosceles triangle, the correct option is 1.
So, the correct answer is “Option 1”.

Note: We can also solve the problem by solving the sides of the triangle taking two sides simultaneously. So, we can get the points of the triangle. From the points we can get the measure of the sides of the triangle and hence determine what type of the triangle it is.