Respuesta :
Answer:
The line passes through the points ( 0 , 4 ) and ( 12 , 0 )
Step-by-step explanation:
Given equation of line as :
y = - [tex]\frac{1}{3}[/tex] x + 4
Let the points through which the line passes are
( [tex]x_1[/tex] , [tex]y_1[/tex] ) , ( [tex]x_2[/tex] , [tex]y_2[/tex] )
Now, for x = 0 , The line equation is y = - [tex]\frac{1}{3}[/tex] × 0 + 4
I.e for x = 0 , y = 0 + 4 = 4
So, ( [tex]x_1[/tex] , [tex]y_1[/tex] ) = ( 0 , 4 )
Again , for y = 0 , The line equation is 0 = - [tex]\frac{1}{3}[/tex] × x + 4
I.e for y = 0 , The line equation is 4 = [tex]\frac{x}{3}[/tex]
Or, for y = 0 , x = 4 × 3 = 12
So , ( [tex]x_2[/tex] , [tex]y_2[/tex] ) = ( 12 , 0 )
Hence The line passes through the points ( 0 , 4 ) and ( 12 , 0 ) Answer
Answer:
The points are (12, 0) & (0, 4).
Step-by-step explanation:
The given equation of line is-
y = -[tex]\frac{1}{3}[/tex] x + 4...............................(i)
After taking L.C.M. both sides (i.e. Multiplying by 3 both sides)-
3 y = -x + 12.................................(ii)
Let x = 0 in equation (ii)-
3 y = 12
y = 4
Hence first point is ( 0, 4).
Now, let y = 0 in equation (ii)-
0 = - x + 12
-12 = - x
x = 12
therefore the points are ( 12, 0).
Hence the required points which passes through the line y = -1/3 x +4 are (0, 4) & ( 12, 0).
