Question

Given three lines L1:x+3y-5=0, L2:3x-ky-1=0 and
L3:5x+2y-12=0, find the value of k such that these three lines do not form a triangle.
(a)k=-9
(b)k= \[-\dfrac{6}{5}\]
(c)k=\[\dfrac{5}{6}\]
(d)k=5

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 equate it to zero. This is because if the triangle area is zero it means that the lines do not form a 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 $C_1, C_2$ and $C_3$ are the cofactors of the elements $c_1, c_2$ and $c3$.
Also remember that if these lines should not form a triangle then the area of this triangle must be 0.
Now let us proceed with this background.
Now comparing with the lines we get,
x+3y-5=0: a1=1, b1=3, c1=-5
3x-ky-1=0: a2=3, b2=-k, c2=-1
5x+2y-12=0: a3=5, b3=2, c3=-12.
Now let us construct the matrix.
\[\left[ \begin{matrix}
   {{a}_{1}} & {{b}_{1}} & {{c}_{1}} \\
   {{a}_{2}} & {{b}_{2}} & {{c}_{2}} \\
   {{a}_{3}} & {{b}_{3}} & {{c}_{3}} \\
\end{matrix} \right]\] = \[\left[ \begin{matrix}
   1 & 3 & -5 \\
   3 & -k & -1 \\
   5 & 2 & -12 \\
\end{matrix} \right]\]
So this is the matrix.
Let us find the values of c1, c2 and c3.
$C_1$ = \[\det \left[ \begin{matrix}
   3 & -k \\
   5 & 2 \\
\end{matrix} \right]\] (we get this by deleting the row and column in which c1 was present)
Therefore, $C_1$=6+5k
$C_2$ = \[\det \left[ \begin{matrix}
   1 & 3 \\
   5 & 2 \\
\end{matrix} \right]\]
Which is $C_2$ = 2-15 = -13
Now $C_3$ = \[\det \left[ \begin{matrix}
   1 & 3 \\
   3 & -k \\
\end{matrix} \right]\]
Which is $C_3$ = -k-9 = -(k+9)
Now let us calculate the determinant of the matrix \[\left| \begin{matrix}
   1 & 3 & -5 \\
   3 & -k & -1 \\
   5 & 2 & -12 \\
\end{matrix} \right|\]
= 1((-k)(-12)-(-1)(2))-3((-12)(3)-(-1)(5))-5((3)(2)-(-k)(5))
Simplifying we get,
=1(12k+2)-3(-36+5)-5(6+5k)
=12k+2-3(-31)-30-25k
=-13k-28+93
=-13k+65
=-13(k-5)
Now that we have got all the values, let us substitute them in our original formula of area and equate it to zero.
\[\dfrac{1}{2|{{C}_{1}}{{C}_{2}}{{C}_{3}}|}{{\left[ \det \left[ \begin{matrix}
   {{a}_{1}} & {{b}_{1}} & {{c}_{1}} \\
   {{a}_{2}} & {{b}_{2}} & {{c}_{2}} \\
   {{a}_{3}} & {{b}_{3}} & {{c}_{3}} \\
\end{matrix} \right] \right]}^{2}}\] = 0
\[\dfrac{1}{2|(6+5k)(-13)(-(k+9))|}{{\left[ \det \left[ \begin{matrix}
   1 & 3 & -5 \\
   3 & -k & -1 \\
   5 & 2 & -12 \\
\end{matrix} \right] \right]}^{2}}\] = 0
\[\dfrac{1}{2(6+5k)(13)(k+9))}{{\left( -13(k-5) \right)}^{2}}\] = 0
\[\dfrac{169}{2(6+5k)(13)(k+9))}{{\left( (k-5) \right)}^{2}}\] = 0
Now if a fraction is 0 its numerator is 0.
Thus, 169(k-5)2 = 0
(k-5)2 = 0
k-5=0
k=5
Thus option(d) is correct

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.