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Answer 1 · 2017-10-16T07:26:35

#y=1/2x#

there are several ways of doing this.

we can use

#y-y_1=m(x-x_1)#

where

#m=" the gradient "=(y_2-y_1)/(x_2-x_1)#

#(x_1,y_1)" is a known coordinate on the line"#

we have

#m=(1-0)/(2-0)=1/2#

#y-y_1=m(x-x_1)#

#=>y-0=1/2(x-0)#

#:.y=1/2x#

or

#2y-x=0#

Answer 2 · 2017-10-16T07:27:58

#:.y=1/2x#

Find the gradient #color(pink)m# of the line by using the formula #(y_2-y_1)/(x_2-x_1)#.

#(1-0)/(2-0)=1/2#

Using the equation #y-y_1=color(pink)m(x-x_1)#, substitute the values of one pair of coordinates.

#y-1=1/2(x-2)#
#:.y=1/2x#