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Line L is perpendicular to line K whose equation is 3y = 4x + 12; Line
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Bunuel
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Line L is perpendicular to line K whose equation is 3y = 4x + 12; Line [#permalink] New post  09 Mar 2016, 12:28
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Line L is perpendicular to line K whose equation is 3y = 4x + 12; Lines L and K intersect at (p, q). Is p + q > 0?

(1) x-intercept of Line L is less than that of Line K
(2) y-intercept of Line L is less than that of Line K
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Nevernevergiveup
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Line L is perpendicular to line K whose equation is 3y = 4x + 12; Line [#permalink] New post  11 Mar 2016, 03:59
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Quote:
Line L is perpendicular to line K whose equation is 3y = 4x + 12; Lines L and K intersect at (p, q). Is p + q > 0?

(1) x-intercept of Line L is less than that of Line K
(2) y-intercept of Line L is less than that of Line K


Line K

Equation \(3y=4x+12\)

Point \((p,q)\) is on line L so \(3q=4p+12\)

X intercept is \(-3\)
Y intercept is \(4\)

Line L

Equation \(y-q=-\frac{3}{4}(x-p)\)
\(3x+4y=3p+4q\)

X intercept is \(\frac{3p+4q}{3}\)
Y intercept is \(\frac{3p+4q}{4}\)

Statement 1: x-intercept of Line L is less than that of Line K

\(\frac{3p+4q}{3}<-3\)

\(3p+4q<-9\)

Sum of p and q i.e., p+q is negative in all cases so sufficient.

Statement 2: y-intercept of Line L is less than that of Line K

\(\frac{3p+4q}{4}<4\)

\(3p+4q<16\)

Sum of p and q i.e., p+q can be both positive and negative hence statement 2 is insufficient.

A for me.
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vardhanindaram
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Re: Line L is perpendicular to line K whose equation is 3y = 4x + 12; Line [#permalink] New post  09 Mar 2016, 22:26
Clue 1 and 2 only suggest that line L can meet K in either quadrant 1 or quadrant 3 as intercepts constraints apply in both the cases.Therefore meeting point p+q can be either positive or negative which can't be decided by given clues.
Hence Answer E
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chetan2u
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Line L is perpendicular to line K whose equation is 3y = 4x + 12; Line [#permalink] New post  09 Mar 2016, 23:50
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vardhanindaram wrote:
Clue 1 and 2 only suggest that line L can meet K in either quadrant 1 or quadrant 3 as intercepts constraints apply in both the cases.Therefore meeting point p+q can be either positive or negative which can't be decided by given clues.
Hence Answer E



Hi,
there is some info in statement 1, which can answer the Q..

Line L is perpendicular to line K whose equation is 3y = 4x + 12; Lines L and K intersect at (p, q). Is p + q > 0?

INFO from this:-


A)the intercept on x axis is (3,0) and y axis is (0,4)..
B) the slope is 4/3, which tells us that y increases at a higher rate than x..

Inference:-


1)Since the Line L is perpendicular to K, If its x-intercept is less than 3, the two lines will intersect above x-axis and if more than 3, it will intersect below X-axis..

lets see the statements


(1) x-intercept of Line L is less than that of Line K
see att fig, p+q will be always negative
a) in Quad III, both p and q will be -ive, so p+q will be NEGAIVE.

Suff..

(2) y-intercept of Line L is less than that of Line K
not suff
insuff..

ans A
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vardhanindaram
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Line L is perpendicular to line K whose equation is 3y = 4x + 12; Line [#permalink] New post  12 Mar 2016, 20:54
chetan2u wrote:
vardhanindaram wrote:
Clue 1 and 2 only suggest that line L can meet K in either quadrant 1 or quadrant 3 as intercepts constraints apply in both the cases.Therefore meeting point p+q can be either positive or negative which can't be decided by given clues.
Hence Answer E



Hi,
there is some info in statement 1, which can answer the Q..

Line L is perpendicular to line K whose equation is 3y = 4x + 12; Lines L and K intersect at (p, q). Is p + q > 0?

INFO from this:-


A)the intercept on x axis is (3,0) and y axis is (0,4)..
B) the slope is 4/3, which tells us that y increases at a higher rate than x..

Inference:-


1)Since the Line L is perpendicular to K, If its x-intercept is less than 3, the two lines will intersect above x-axis and if more than 3, it will intersect below X-axis..

lets see the statements


(1) x-intercept of Line L is less than that of Line K
see att fig, p+q will be always negative
a) in Quad III, both p and q will be -ive, so p+q will be NEGAIVE.

Suff..

(2) y-intercept of Line L is less than that of Line K
not suff
insuff..

ans A


The line that you drew in the picture is wrong.It has to meet x axis at -3.
And I would like someone to clarify if intercept is said to be less than -3,does that mean intercept of line should be between 3 and -3 considering the definition of intercept to be distance between origin and point where line meets axis.This is if we ignore the sign as it's the convention to denote the direction.
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chetan2u
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Line L is perpendicular to line K whose equation is 3y = 4x + 12; Line [#permalink] New post  12 Mar 2016, 21:09
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vardhanindaram wrote:
chetan2u wrote:
vardhanindaram wrote:
Clue 1 and 2 only suggest that line L can meet K in either quadrant 1 or quadrant 3 as intercepts constraints apply in both the cases.Therefore meeting point p+q can be either positive or negative which can't be decided by given clues.
Hence Answer E



Hi,
there is some info in statement 1, which can answer the Q..

Line L is perpendicular to line K whose equation is 3y = 4x + 12; Lines L and K intersect at (p, q). Is p + q > 0?

INFO from this:-


A)the intercept on x axis is (3,0) and y axis is (0,4)..
B) the slope is 4/3, which tells us that y increases at a higher rate than x..

Inference:-


1)Since the Line L is perpendicular to K, If its x-intercept is less than 3, the two lines will intersect above x-axis and if more than 3, it will intersect below X-axis..

lets see the statements


(1) x-intercept of Line L is less than that of Line K
see att fig, p+q will be always negative
a) in Quad III, both p and q will be -ive, so p+q will be NEGAIVE.

Suff..


(2) y-intercept of Line L is less than that of Line K
not suff
insuff..

ans A


The line that you drew in the picture is wrong.It has to meet x axis at -3.
And I would like someone to clarify if intercept is said to be less than -3,does that mean intercept of line should be between 3 and -3 considering the definition of intercept to be distance between origin and point where line meets axis.This is if we ignore the sign as it's the convention to denote the direction.


Hi,
Thanks for pointing out..
Firstly less than -3 should mean <-3 in literal sense..
now the point of intersection would be in III quad, where both x and y are -ive, so the SUM will always be -ive..
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rohit8865
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Line L is perpendicular to line K whose equation is 3y = 4x + 12; Line [#permalink] New post  Updated on: 23 Oct 2016, 07:09
Bunuel wrote:
Line L is perpendicular to line K whose equation is 3y = 4x + 12; Lines L and K intersect at (p, q). Is p + q > 0?

(1) x-intercept of Line L is less than that of Line K
(2) y-intercept of Line L is less than that of Line K


Please refer fig. with above solution for better understanding
Attachments

coordinate (1).png
coordinate (1).png [ 12.68 KiB | Viewed 5446 times ]


Originally posted by rohit8865 on 10 Oct 2016, 09:21.
Last edited by rohit8865 on 23 Oct 2016, 07:09, edited 1 time in total.
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kplusprinja1988
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Re: Line L is perpendicular to line K whose equation is 3y = 4x + 12; Line [#permalink] New post  10 Oct 2016, 09:52
Nevernevergiveup wrote:
Quote:
Line L is perpendicular to line K whose equation is 3y = 4x + 12; Lines L and K intersect at (p, q). Is p + q > 0?

(1) x-intercept of Line L is less than that of Line K
(2) y-intercept of Line L is less than that of Line K


Line K

Equation \(3y=4x+12\)

Point \((p,q)\) is on line L so \(3q=4p+12\)

X intercept is \(-3\)
Y intercept is \(4\)

Line L

Equation \(y-q=-\frac{3}{4}(x-p)\)
\(3x+4y=3p+4q\)

X intercept is \(\frac{3p+4q}{3}\)
Y intercept is \(\frac{3p+4q}{4}\)

Statement 1: x-intercept of Line L is less than that of Line K

\(\frac{3p+4q}{3}<-3\)

\(3p+4q<-9\)

Sum of p and q i.e., p+q is negative in all cases so sufficient.

Statement 2: y-intercept of Line L is less than that of Line K

\(\frac{3p+4q}{4}<4\)

\(3p+4q<16\)

Sum of p and q i.e., p+q can be both positive and negative hence statement 2 is insufficient.

A for me.



can you explain how did you deduce 3p+4q<-9[/m]

Sum of p and q i.e., p+q is negative in all cases so sufficient.
and
3p+4q<16[/m]

Sum of p and q i.e., p+q can be both positive and negative hence statement 2 is insufficient.
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Crytiocanalyst
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Re: Line L is perpendicular to line K whose equation is 3y = 4x + 12; Line [#permalink] New post  18 Jul 2021, 10:44
First of let us pen down the equation of the lines
k-3y = 4x + 12(given)
=>y=-3/4x+c since it's perpendicular to k
now we have to figure out whether p+q point of intersection of k and l that is p+q>0

state 1 provides :
c*4/3<-3 that is let us assume c=-10/4 <-9/4
gives us p+q<0 when substitued in L
sufficient

state 2 provides :
in a similar method of substitution provides c<-3
however when finding out for p and q provides both p+q>0
and p+q<0
hence Insuff

IMO A
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Re: Line L is perpendicular to line K whose equation is 3y = 4x + 12; Line [#permalink]
18 Jul 2021, 10:44
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