Question

Find the value of p. If the following lines are concurrent 4x-3y-7=0,2x+py+2=0,6x+5y-1=0​

Answer 1

Given:

The following lines are concurrent 4x-3y-7=0,2x+py+2=0,6x+5y-1=0​

To find:

Find the value of p.

Solution:

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,

$\det \begin{pmatrix}4&-3&-7\\ 2&p&2\\ 6&5&-1\end{pmatrix}\\=4\cdot \det \begin{pmatrix}p&2\\ 5&-1\end{pmatrix}-\left(-3\right)\det \begin{pmatrix}2&2\\ 6&-1\end{pmatrix}-7\cdot \det \begin{pmatrix}2&p\\ 6&5\end{pmatrix}\\=4\left(-p-10\right)-\left(-3\right)\left(-14\right)-7\left(10-6p\right)$

= 38p - 152

⇒ 38p - 152 = 0

38p = 152

p = 152/38

p = 4

Therefore, the value of p is 4.

Answer 2

Step-by-step explanation:

The conditions of lines to be concurrent is that, the determinant of these lines forming a matrix should be equal to zero.....Therefore the value of p is 4