Question

Given three lines $L1: 2x+2y-4=0, L2: 3x-ky-3=0$ and
$L3: 6x+3y-9=0$, find the value of k such that these three lines form a triangle.
(a). $k=-3$
(b). \[k= -\dfrac{3}{2}\]
(c). \[k= 5\]
(d). \[k=0\]

Answer 1

Hint: The hint here is that if the lines will form a triangle, the triangle will have a non-zero area. So we will find the area of the triangle formed by the three given lines first and then we will make sure that the area is not zero by verifying the options given. This is because if the triangle area is not zero it means that the lines form a well defined triangle.

Complete step-by-step solution:
Let us solve this question using the concept of finding area using determinants. There is a very nice formula if you know to find the area of triangle formed by three sides which is
\[\dfrac{1}{2|{{C}_{1}}{{C}_{2}}{{C}_{3}}|}{{\left( \det \left( \begin{matrix}
   {{a}_{1}} & {{b}_{1}} & {{c}_{1}} \\
   {{a}_{2}} & {{b}_{2}} & {{c}_{2}} \\
   {{a}_{3}} & {{b}_{3}} & {{c}_{3}} \\
\end{matrix} \right) \right)}^{2}}\]
Where C1, C2 and C3 are the cofactors of the elements c1, c2 and c3.
Also remember that if these lines form a triangle then the area of this triangle must not be 0.
Also this problem needs a concept of fractions in which we need to determine which values will make the fraction undefined or zero.
Now let us proceed with this background.
Now comparing with the lines we get,
$2x+2y-4=0: a1=2, b1=2, c1=-4$
$3x-ky-3=0: a2=3, b2=-k, c2=-3$
$6x+3y-9=0: a3=6, b3=3, c3=-9.$
Now let us construct the matrix.
\[\left( \begin{matrix}
   {{a}_{1}} & {{b}_{1}} & {{c}_{1}} \\
   {{a}_{2}} & {{b}_{2}} & {{c}_{2}} \\
   {{a}_{3}} & {{b}_{3}} & {{c}_{3}} \\
\end{matrix} \right)=\left( \begin{matrix}
   2 & 2 & -4 \\
   3 & -k & -3 \\
   6 & 3 & -9 \\
\end{matrix} \right)\]
So this is the matrix.
Let us find the values of c1, c2 and c3.
C1 = \[\det \left( \begin{matrix}
   3 & -k \\
   6 & 3 \\
\end{matrix} \right)\] (we get this by deleting the row and column in which c1 was present)
Therefore, $C1=9+6k$
C2 = \[\det \left( \begin{matrix}
   2 & 2 \\
   6 & 3 \\
\end{matrix} \right)\]
Which is $C2 = 6-12 = -6$
Now $C3$ = \[\det \left( \begin{matrix}
   2 & 2 \\
   3 & -k \\
\end{matrix} \right)\]
Which is $C3 = -2k-6 = -\left(2k+6 \right)$
Now let us calculate the determinant of the matrix \[\left( \begin{matrix}
   2 & 2 & -4 \\
   3 & -k & -3 \\
   6 & 3 & -9 \\
\end{matrix} \right)\]
$= 2((-k)(-9)-(-3)(3))-2((-9)(3)-(-3)(6))-4((3)(3)-(-k)(6))$
Simplifying we get,
$=2(9k+9)-2(-27+18)-4(9+6k)$
$=18k+18-2(-9)-36-30k$
$=18k+18+18-36-24k$
$= -6k$
Now that we have got all the values, let us substitute them in our original formula of area and equate it to zero.
\[\dfrac{1}{2|{{C}_{1}}{{C}_{2}}{{C}_{3}}|}{{\left( \det \left( \begin{matrix}
   {{a}_{1}} & {{b}_{1}} & {{c}_{1}} \\
   {{a}_{2}} & {{b}_{2}} & {{c}_{2}} \\
   {{a}_{3}} & {{b}_{3}} & {{c}_{3}} \\
\end{matrix} \right) \right)}^{2}}\ne 0\]
\[\dfrac{1}{2|(9+6k)(-6)(-(2k+6))|}{{\left( \det \left( \begin{matrix}
   2 & 2 & -4 \\
   3 & -k & -3 \\
   6 & 3 & -9 \\
\end{matrix} \right) \right)}^{2}}\ne 0\]
\[\dfrac{1}{2(9+6k)(6)(-2k-6))}{{\left( -6k \right)}^{2}}\ne 0\]
\[\dfrac{36}{2(9+6k)(6)(2k+6))}{{\left( k \right)}^{2}}\ne 0\]
\[\dfrac{3}{(9+6k)(2k+6))}{{\left( k \right)}^{2}}\ne 0\]
Now let us verify the options one by one and look that the fraction above does not become zero.
Option(a) says that $k=-3$. But substituting $k=-3$ makes the $(2k+6)$ term 0 which makes the entire fraction undefined. So option(a) is not correct.
Option(b) says that \[k= -\dfrac{3}{2}\] . But substituting this value makes the $(9+6k)$ term become zero which makes the entire fraction again undefined. So option(b) is also not correct.
Option(c) says that \[k= 5 \] . This substitution does not make the fraction 0 or does not make it undefined. So this is the correct option.
Let us not stop here. What if this is a “multiple correct answers” type of question. So better to verify all the options.
Option(d) says that $k=0$. But this makes the fraction 0 which we do not want. Thus option(d) is wrong.
Thus option(c) is the correct option.

Note: You should be very careful about the square term in the area of the triangle formula. The formula says first calculate the determinant of the matrix and then square the value.
Most of the students get confused and first find the square of the matrix and then find its determinant. It’s a completely wrong method which not only will give you a wrong answer, but also eat a lot of your time.
Also remember that if these lines form a triangle then the area of this triangle must not be 0.