How do you find a standard form equation for the line with point (1, 1) and has a y-intercept?

2 Answers
Apr 28, 2017

see explanation

Explanation:

Standard form of an equation involves two intercepts, x and y intercepts...

...but you only have one intercept, and it's not complete.
Is this all the information you have?

Apr 28, 2017

#y=x#

The y-intercept #->(x,y)=(0,0)#

Explanation:

#color(blue)("Initial reaction")#

Initially, due to the lack of information it is tempting to write:

The gradient is such that you read left to right on the x-axis> So to force this to work we must have:

For the gradient set #x_g>1#
For the gradient set #y_g>1# where #y_g=f(x_g) larr" some function of "x#

Under these conditions we can write:
Let the gradient be #m=(y_g-1)/(x_g-1)#

Let the y-intercept be #c#

Then we have #y=mx+c#
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("However")#

We are given that the line passes through the point (1,1)

Set the point as #P_1->(x_1,y_1)=(1,1)#

So we have the answer of 1 when #x=1#

Thus the standard form of #y=mx+c# can only work with the following two conditions:

#m=1 and c=0#

Note that #c# is the y-intercept

Thus as the y-intercept is at #x=0# we have:

#y=1(0)+0#

#=>y=0->P_2->(x_2,y_2)=(0,0)#

Thus #y=mx+c" becomes "y=x#