You got the answer correct

General formula y = mx+c (m is slope)

Passes through (0,0). Put in the formula c=0

Thus y = mx

passes through (a,b) then b = ma

1. slope is -ve
Then b = -m*a then (a,b) can lie in 2nd or 4th quadrant.
if 2nd then b +ve
if 4th then b -ve

Not sufficient.

2. a<b
we cannot say anything as this can happen in all of the quadrants. Not sufficient.

Combine both only 2nd qudarant will satisfy as a will be -ve and b will be +ve. Thus sufficient.

PS: Slope negative and line must through origin then the line must be in 2nd to 4th quadrant,

See below 3 lines are there in the picture.

1. Red: Angle subtended by this line with X axis is less than 90. Slope is +ve. (Remember trigonometry? TAN(angle) is the slope)
2. Blue: Angle subtended by this line with X axis is equal to 90. Slope is infinite(cannot be determined).
3. Green: Angle subtended by this line with X axis is greater than 90. Slope is -ve.

@Shovan: Thanks for the awesome explanation.
You really attack the problem right at the fundamentals. I like the approach.
Thanks for sharing the diagram.

euro wrote:@Shovan: Thanks for the awesome explanation.
You really attack the problem right at the fundamentals. I like the approach.
Thanks for sharing the diagram.

Thanks!! Much appreciated. Please pray for me I have my D-Day on 2nd Nov

Hi,

For C (both together) :

I get that 2nd quadrant satisfies the conditions but why not quad 4?
We could have a line with negative slope that passes through origin and in the 4th quadrant a point -4,-2 where a<b
but b is not greater than 0..

Rezinka wrote:Hi,

For C (both together) :

I get that 2nd quadrant satisfies the conditions but why not quad 4?
We could have a line with negative slope that passes through origin and in the 4th quadrant a point -4,-2 where a<b
but b is not greater than 0..

@Rezinka:
In 4th quad, values of x are always positive. Point -4, -2 will lie in the 3rd quad and not in the 4th.

Oh..!!
Silly me..!!
Thanks