xcusemeplz2009 wrote:i was able to solve it graphically , intrested in how to find the slope non graphically.
and pls suggest ways to find out slope and
how to know when is it negative or positive if drawn in a graph??
Hi xcusemeplz2009,
reading from left to right, if the line falls, the slope is negative. If the line rises, the slope is positive. Slope = rise/run= change in y/change in x. When two lines are parallel, their slopes are equal. When two lines are perpendicular their slopes are negative reciprocals.
In this question, honestly, if you understand that the product of the slopes will be negative when one line rises and one line falls, then a graphical approach is by far the quickest way to answer this question. Because we are asked about the sign of the product of the slopes and because the statements provide info about the signs of intercepts, the only real significance of the intersection being 4,3 is that it is in the first quadrant.
But if you really want to do it algebraically:
Let's use "1" to denote line L, and "2" to denote line K.
Then:
Line L: y1 = m1x1 + b1
Line K: y2 = m2x2 +b2
in which m is slope and b is y intercept.
Because they meet at (4,3), we can use "3" as the y coordinate for each line and equate the line equations of each line.
subbing in "4" for the x coordinate:
4m1 + b1 = 4m2 + b2
The question is asking whether m1*m2 < 0
1) The product of the x-intercepts of line L and K is positive.
At the x intercept, the y coordinate equals zero. So, we can sub zero into the line equation of each line to have an expression for each line's x intercept:
Line L:
0 = m1x1 + b1;
X intercept = -b1/m1
Likewise, the x intercept of line K is: -b2/m2
This statement tells us that the product of these x intercepts is positive:
(-b1/m1)*(-b2/m2) >0
or:
b1*b2/m1*m2>0
So, m1*m2 need not be positive. If the numerator (b1*b2) is positive, then m1*m2 is positive. But if the numerator is negative, then, in order to satisfy the statement, m1*m2 is also negative.
In other words, if b1*b2 is pos, then m1*m2 is pos. But if b1*b2 is neg, then so is m1*m2.
Because we can get both a yes and no answer, this statement is not sufficient.
2) The product of the y-intercepts of line L and k is negative.
Knowing that the product of the y intercepts is negative is clearly insufficient. If b1*b2<0, all we know is that one line's y intercept is negative, and the other positive.
combo:
In combination b1*b2<0, and:
b1*b2/m1*m2 >0
This is only possible if m1*m2 <0.
Combined, the answer is yes.
Choose C.
To understand some of the qualitative aspects of coordinate geomtry better, this post of mine might help:
https://www.beatthegmat.com/gmat-prep-ge ... 14275.html
