Find the equation, in standard form, of the line passing through the points (2,-3) and (4,2).
let us do this step-by-step okay? I will give you hints and you the rest
first, solve for the slope
ok
the slope is 5/2, right?
\[slope=m=\frac{ \Delta y }{ \Delta x }=\frac{ rise }{ run }=\frac{ y_2 - y_1 }{ x_2 - x_1 }\]
Ok, i did that. I did 2-(-3)/ 4-2 and i got 5/2
now that you have the slope, use the point-slope form equation \[y - y_1 = m(x-x_1)\]
ooh
go head, plug in the numbers
y-2=5/2(x-4)?
use distributive property to multiply the slope into x and -4
y-2=5/2x-10
not 5/2x it is 5x/2 be very careful
ah
add 2 to both sides?
\[\frac{ 5 }{ 2 } \times \frac{ x }{ 1 } = \frac{ 5x }{ 2 }\]
then solve for y you can achieve this by isolating y and in your problem you need to add 2 on both sides of the equation (left and and right hand sides)
and that's the answer in standard form? y=5x/2-8?
nope that is the slope form
standard form is \[Ax+By=C\]
so would it be 5x+2y=8?
how so?
because the 5 is next to x, the -8 is constant and 2 is what's left i guess? and i'm not sure if they can be negative
first, rearrange it in the format of Ax + By = C the coefficient of x is 5/2 the coefficient of y is 1 and C = 8 rearranging \[y = \frac{ 5 }{ 2 }x - 8\] add 8 to both sides subtract y to both sides then make sure that everything is in whole number
always make sure that A is a positive integer
ok, so is it 5x-2y=16?
yes
yay, thank you so much :)!
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