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 "#
#"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"#
