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Fig, I am not very good at coordinate geometry. Could you please explain this, especially the part written in red? A negative slope for a line to pass by the cadran II (y>0 and x<0).
We have : y = -1/6*x + b... If x < 6*b and x <0> 0 and we are in the cadran II.
I prefer to draw an XY plan ...
All lines sharing a similar slope are parallel to one another and called a family of parallel lines. The most simple line of a family is of a kind : y = a*x. Here, we have y = -1/6 * x... Look at the Fig 1... The line passes by the cadran II. Imagine any parallel line to it, it must pass by the cadran II
And, to respond to your question, we can rethink of the definition of the cadran II: y>0 and x<0... Let see if x could be negative when y is postive.
<=> -1/6*x + b > 0 as y = -1/6*x + b, the equation of the line
<=> 1/6*x < b
<=> x < 6 * b >>>>> So, yes, x could be negative when y > 0.
Fig1_Family of - x div 6.gif [ 2.94 KiB | Viewed 838 times ]
Q: In a rectangular coordinate system, does the line k intersect quadrant II?
(1) the slope of k is -1/6 this means equation of line is y=-x/6+c where c is y intercept if y=0 then x=6c means x can be positive or negative for y=0 so it is not sufficient(2) y-intercept of k is -6
similary 2 alone is not sufficient
but if we combine 1 and 2, then line equation is y=-x/6-6 points of this line are (-36,0) and(0,-6) which intercepts quadrant III not II
so C is the answer for me
If you carefully look at the stem- we have to find out whether the line with slope -1/6 passes thru the second quadrnt.
the equation of the line will be y =(-1/6)x + c where c =y intercept (when x=0)
we don't know about c. The possible values of c are zero, -ve and +ve numbers. For all values of c, the line passes through the quad II.
Hence statement 1 is sufficient.
Statement 2 is insuff as we don't know the slope of the line so not possible to know where the line will pass through from.