find the value of p, if the following lines are 4x-3y-7=0,2x+py+2=0,6x+5y-1=0
Answers
Step-by-step explanation:
From given, we have,
The following lines are concurrent 4x - 3y - 7 = 0, 2x + py + 2 = 0, 6x + 5y - 1 = 0
The condition for lines to be concurrent is that, the determinant of these lines forming a matrix should be equal to 0.
So, we have,
\begin{gathered}\det \begin{pmatrix}4&-3&-7\\ 2&p&2\\ 6&5&-1\end{pmatrix}\\=4\cdot \det \begin{pmatrix}p&2\\ 5&-1\end{pmatrix}-\left(-3\right)\det \begin{pmatrix}2&2\\ 6&-1\end{pmatrix}-7\cdot \det \begin{pmatrix}2&p\\ 6&5\end{pmatrix}\\=4\left(-p-10\right)-\left(-3\right)\left(-14\right)-7\left(10-6p\right)\end{gathered}
det
⎝
⎛
4
2
6
−3
p
5
−7
2
−1
⎠
⎞
=4⋅det(
p
5
2
−1
)−(−3)det(
2
6
2
−1
)−7⋅det(
2
6
p
5
)
=4(−p−10)−(−3)(−14)−7(10−6p)
= 38p - 152
⇒ 38p - 152 = 0
38p = 152
p = 152/38
p = 4
Therefore