Question

The value of p for which the lines 3x -(p-1)y= 7;(p+1)x -5y =8 are parallel

Pleaseeee help

Answer 1

Answer:

Given equation,

3x+y+5=0,6x+2y+p=0

If the equations a

1 x+b 1 y+c 1=0

then 2 x+b 2 y+c 2 = represent parallel lines,

➡ refer the attachment....

hence p∈R−{10}, to represent parallel lines.

Answer 2

$\huge \tt \red{SOLUTION} :$

For parallel lines, the condition is,

$\implies \boxed{ \dfrac{a1}{a2} = \dfrac{b1}{b2} ≠\dfrac{c1}{c2} }$

Here, a1 = 3 , b1 = - ( p-1) , c1 = 7

a2 = (p+ 1) , b2 = -5 , c2 = 8

So,

$\dfrac{3}{p + 1} = \dfrac{ - p + 1}{ - 5} ≠ \dfrac{7}{8}$

Considering,

$\dfrac{3}{p + 1} = \dfrac{ - p + 1}{ - 5}$

$- {p}^{2} + p - p + 1 = - 15$

$- {p}^{2} + 1 = - 15$

${p}^{2} = 16$

$\implies \large \red{\boxed{p = 4}}$

______________________________

therefore, p = 4