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805+ Level|   Geometry|      
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Are lines p (with slope m) and q (with slope n) perpendicular to each 15 Apr 2019, 03:00
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Are lines p (with slope m) and q (with slope n) perpendicular to each other?

(1) \(m+2=2n\)
(2) \(m+2=n\)
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Are lines p (with slope m) and q (with slope n) perpendicular to each 19 Apr 2019, 02:20
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I would really appreciate if someone can explain me this. I don't understand how the option 1) alone is sufficient.

Thanks in advance
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Are lines p (with slope m) and q (with slope n) perpendicular to each 26 Apr 2019, 05:25
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Can someone please post an explanation why statement 1 is sufficient?
TIA!
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Are lines p (with slope m) and q (with slope n) perpendicular to each 26 Apr 2019, 08:59
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For the two lines to be perpendicular, the product of their slopes must equal -1.
(1) m + 2 = 2n means m = 2n - 2. Therefore, mn= -1 if and only if (2n - 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.

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Are lines p (with slope m) and q (with slope n) perpendicular to each 09 Jun 2021, 21:58
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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.
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Are lines p (with slope m) and q (with slope n) perpendicular to each 09 Jun 2021, 22:14
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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
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