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by tanviet » Mon Mar 15, 2010 4:16 am
k and l cut each other at the point (4,3), is the products of their slope is nagative

1, product of their x-intercepts is positive

2, product of their y-intercepts is negative
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by girish3131 » Mon Mar 15, 2010 4:46 am
Good ques...

acc to me .. Ans is E

plz specify OA

ta
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by rockeyb » Mon Mar 15, 2010 4:49 am
IMO E .
"Know thyself" and "Nothing in excess"
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by tanviet » Thu Mar 18, 2010 11:28 pm
how to get answer. I do not have oa
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by neoreaves » Fri Mar 19, 2010 6:50 am
I think OA is C

we know that y = mx + c

for line L : y = m1x + c1
for line M : y = m2x + c2

What we need to know is m1m2<0 ?

1) x intercept = -c/m ( Just put y = 0 and solve)

(-c1/m1)(-c2/m2) > 0

Insufficient
2)c1*c2<0 insufficient

C) Combining inequalities in 1) and 2)

m1m2<0

Thus answer should be C
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by kstv » Fri Mar 19, 2010 7:23 pm
The lines intersect at (4,3) so the lines are in the I st Quadrant
The slope of a line is = y intercept/x intercept
Let us asssume for line 1 the x and y intercept are x1, y1 and line 2 x2,y2
given x1*x2 is +ve and y1*y2 is +ve taking 1 & 2 together
so either x1 and x2 are both +ve or both -ve same for y1 and y2
but since they both pass through (4,3) they have to be +ve and product of slope is +ve

x1*x2 is +ve Given in Q stem 1 and we do not consider that y1*y2 is +ve
y1*y2 can be + or - already discussed when it is +ve
if it is -ve y1 & y2 have diff signs say y1 is +ve y2 is -ve
slope of 1st line y1/x1 both are +ve so slope is +ve slope of 2nd line y2/x2 y2 is -ve
so product of their slope is -ve
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by outreach » Sat Mar 20, 2010 12:35 am
there are three scenarios at which the 2 lines will intersect the x and y axis after passing through pt P
-line can intersect each axis at (+,+). in this case the slope is negative
-line can intersect each axis at (-,+) or (+,-). in both the cases the slope will be positive
- (-,-) is not possible bcz line passed thr pt (4,3)

opt 1: product of their x-intercepts is positive

in this case the 2 lines can intersect the x and y axis at
- (+,+) and (+,+) => products of their slope will be positive
- (+,+) and (+,-) => products of their slope will be negative

insufficient


opt 2:, product of their y-intercepts is negative
- (+,+) and (+,-) => products of their slope will be negative
- (-,+) and (+,-) => products of their slope will be positive

insufficient

opt 1 and opt 2 combined
(+,+) and (+,-) => products of their slope will be negative

hence C
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by tanviet » Sun Mar 21, 2010 8:30 pm
thank you outstreach

your solution is comprehensive
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by harshavardhanc » Mon Mar 22, 2010 2:16 am
duongthang wrote:k and l cut each other at the point (4,3), is the products of their slope is nagative

1, product of their x-intercepts is positive

2, product of their y-intercepts is negative
i don't think the point really matters here.

we know that the slope is y-intercept/X-intercept.

We need to find out if (y1/x1) * (y2 /X2) or ( y1*y2) / (x1*x2) is negative.

Statement 1 says (x1 * x2) is +ve. We don't know what is the sign of (y1 * y2), so we can't say that ( y1*y2) / (x1*x2) will be -ve.

Hence, insufficient.


Statement 2 says (y1 * y2) is -ve. We don't know what is the sign of (x1 * x2), so we can't say that ( y1*y2) / (x1*x2) will be -ve.

Hence, insufficient.

But, combining these two statements we can definitely say that ( y1*y2) / (x1*x2) will be -ve.


Hence, IMO C.
Regards,
Harsha
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