If (p + 1) x + y = 3 and 3y – px + (x – 4) = 0. Are perpendicular to each other, find the value of p.
Answers
Answer:
The value of p is 2. If tho helps ... mark as brainliest please.
Step-by-step explanation:
From the first equation
(p+1)x + y = 3
make y the subject of the formula
y = -(p+1)x + 3
from equation of a straight line
y = mx + c
; m = -(p+1)............ 1
Second equation
3y - (p-1)x=4
make y the subject of the formula
3y = (p-1)x + 4
y = (p-1)x/3 + 4/3
comparing equation of a straight line
y = mx + c
m = (p-1)/3........... 2
since the two equations are perpendicular
m2 = -1/(m1)
\begin{gathered} \frac{p - 1}{3} = \: \frac{ - 1}{ - (p + 1)} \\ cross \: multiplication \\ (p - 1)(p + 1) \: = \: 3 \: \: negative \: v \\ cancelled \: out. \\ by \: expansion.... \\ {p}^{2} - 1 = 3 \\ {p}^{2} = 4 \\ p = \sqrt{4} \\ p = 2\end{gathered}
3
p−1
=
−(p+1)
−1
crossmultiplication
(p−1)(p+1)=3negativev
cancelledout.
byexpansion....
p
2
−1=3
p
2
=4
p=
4
p=2