It is known that (2, - 1) is a solution of the equation kx-y = 0, then the quadrant that the image of a straight line y = KX + 1 does not pass through is A: First quadrant B: second quadrant C: third quadrant D: fourth quadrant Given that Y1 = - x + 2, y2 = x + 4, when x takes what value, Y1 > Y2? | Math enthusiast | Mathematical art

It is known that (2, - 1) is a solution of the equation kx-y = 0, then the quadrant that the image of a straight line y = KX + 1 does not pass through is A: First quadrant B: second quadrant C: third quadrant D: fourth quadrant Given that Y1 = - x + 2, y2 = x + 4, when x takes what value, Y1 > Y2?

Substituting (2, - 1) into the equation:
2K + 1 = 0, k = - 0.5, so the function is y = - 0.5x + 1,
Because when k is negative, the image passes through 2,4 quadrants
And because B = 1, which is greater than zero, it goes through quadrant 1
So three quadrants, C
Draw the image of Y1 and Y2, when Y1 line is higher than Y2, that is, Y1 is greater than Y2

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