Question

How do you find a standard form equation for the line with (1,3) and (5,9)?

Answer 1

`y = 3/2 x + 3/2`

Explanation:

We must use our knowledge of line graphs:

The general equation is: `y = mx +c`

Where `m` is our gradient...

To find `m`:

`m ="change in y" /"change in x" = (Deltay) / (Delta x)`

`=> m = (9-3)/(5-1) = 6/4 = 3/2`

`=> y= 3/2 x + c`

Now to find `c`:

We can just substitute one of the points in:

`3 = (3/2*1) +c`

Subtracting `3/2` from both sides...

`3-3/2 = c`

`c = 3/2`

`=> y = 3/2x + 3/2`

`=> y = 3/2 ( x+1 )` if you understand factorisation...

Answer 2

`3x-2y=-3`

Explanation:

`"the equation of a line in "color(blue)"standard form"` is.

`color(red)(bar(ul(|color(white)(2/2)color(black)(Ax+By=C)color(white)(2/2)|)))`

`"where A is a positive integer and B, C are integers"`

`"the equation of a line in "color(blue)"slope-intercept form"` is.

`•color(white)(x)y=mx+b`

`"where m is the slope and b the y-intercept"`

`"to calculate m use the "color(blue)"gradient formula"`

`color(red)(bar(ul(|color(white)(2/2)color(black)(m=(y_2-y_1)/(x_2-x_1))color(white)(2/2)|)))`

`"let "(x_1,y_1)=(1,3)" and "(x_2,y_2)=(5.9)`

`rArrm=(9-3)/(5-1)=6/4=3/2`

`rArry=3/2x+blarrcolor(blue)"is the partial equation"`

`"to find b use either of the two given points"`

`"substituting "(1,3)" into the partial equation"`

`3=3/2+brArrb=3/2`

`rArry=3/2x+3/2larrcolor(red)"in slope-intercept form"`

`"multiply through by 2"`

`2y=3x+3`

`rArr3x-2y=-3larrcolor(red)"in standard form"`