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