How to find the image of the straight line?
1 Answer
Feb 2, 2018
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"