Are lines p (with slope m) and q (with slope n) perpendicular to each

Question

Are lines p (with slope m) and q (with slope n) perpendicular to each other?

(1)  m+2=2n 
(2)  m+2=n

vedavyas003 · Answer

I would really appreciate if someone can explain me this. I don't understand how the option 1) alone is sufficient.

Thanks in advance

neha283 · Answer

Can someone please post an explanation why statement 1 is sufficient?
TIA!

ndcuong · Answer

For the two lines to be perpendicular, the product of their slopes must equal -1. 
(1)  m  + 2 = 2 n  means  m  = 2 n  - 2. Therefore,  m  n = -1 if and only if (2 n  - 2) n  = -1. But the latter equation is a quadratic equation with discriminant less than zero and so has no real solutions. That implies that the product of the two slopes cannot take value -1, meaning the lines are never perpendicular. Thus (1) alone is sufficient.
(2) if  m  + 2 =  n , then  mn  =  m ( m  + 2). The equation  m ( m  + 2) = -1 can be rewritten as ( m  + 1) ^2 = 0, which has one solution  m  = -1. So if  m  = -1, the two lines are perpendicular. But it can be seen easily that the expression  m ( m  + 2) can also take many other values other than -1. Therefore, (2) alone is not sufficient.

vedavyas003 ,  neha283

sajjad1995 · Answer

For lines p and q to be perpendicular with each other,   m*n=-1

Now, evaluate statement -1,

m+2=2n
put m=-1/n
Therefore, (-1/n)+2=2n
On simplifying above equation,
We get 2 n^2 -2n+1=0.
By comparing above quadratic equation with standard quadratic equation a x^2 +bx+c=0, 
We have a=2, b=-2, c=1
Discriminant, D =  b^2 -4*a*c
 =  2^2 -4*2*1 
 = 4-8
 = -4  (D<0) 
 Hence above equation has no real roots or has imaginary roots.
Therefore , we can say there is no real values exist for m and n for lines p and q to be perpendicular with each other. 
We have a  DEFINITE "NO"  as a answer.
 Hence, Statement -1 is sufficient.

Now evaluate statement -2;
Statement -2 : m+2=n
We will use counter example approach

Case-1: Lines are not perpendicular, 
For m=1, we have n=3 (infinite possible values are there)

Case-2: Lines are not perpendicular, 
mn=-1
m=-1 and n=1 (satisfy condition for lines to be perpendicular )
-1+2=1
1=1 (satisfies m+2=n)

Therefore, with statement -2 , we have no definite answer.
Hence, Statement-2 is NOT sufficient.

Hence,  Statement-1 alone is sufficient  and  Statement-2 alone is not sufficient .

Choice A is the answer   .

Ved3 · Answer

i) Given m + 2 = 2n
So m - 2n = -2
 (m - 2n)^2 = 4
 m^2 +4n^2 -4mn = 4
 m^2 + 4n^2 -4 = 4mn
We know that for the lines to be perpendicular mn should be equal to -1 so putting that in the equation we get
 m^2 + 4n^2 -4 = -4
 m^2 + 4n^2 = 0
We know that sum of the square of two non-zero real numbers can not be 0.
For m = 0 and n = not defined or vice versa is not possible due to the equation, so the lines are not perpendicular
ii) Given m + 2 = n
 So mn = m(m+2)
 mn = m^2 + 2m
 This is perpendicular only for 1 value of m i.e m = -1
So we can't say

There for i) is sufficient but ii) is not

Are lines p (with slope m) and q (with slope n) perpendicular to each

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Are lines p (with slope m) and q (with slope n) perpendicular to each

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