How to find the image of the straight line?

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1 Answer
Feb 2, 2018

y=-1/6x-11/2y=16x112

Explanation:

"any point on the line "y=-2x+6" will have "any point on the line y=2x+6 will have
"coordinates of the form "(k,-2k+6)coordinates of the form (k,2k+6)

rArr((x'),(y'))=((0,-3),(1,0))((k),(-2k+6))+((-3),(2))

color(white)(xxxxxxx)=((6k-18),(k))+((-3),(2))

rArrx'=6k-21" and "y'=k+2

"eliminating k from these 2 equations gives"

6y'+x'=-33

"the line "6y+x+33=0" is the image of "y=-2x+6

"or "y=-1/6x-11/2larrcolor(blue)"in slope-intercept form"