How do you find a standard form equation for the line with (0,3),(-5,0)?

Question

Julia Adams · Accepted Answer

#3x-5y=-5#
#"Or "#
#y=3/5 x+3# #y_2-y)/(x_2-x)= #(y-y_1)/(x-x_1)# #B(x_2,y_2)=(-5,0)# and #A(x_1,y_1)=(0,3)# #(y-3) / (x-0) = 0-y / (-5-x) # (-y)/(-5-x)# = #(y-3)/(x) #-x y=(y-3)(-5-x)# 5 y-y x+15+3x# = #-x y # "rearrange" # #15#= #3x-5y-cancel(y x)+cancel(y x) #3x-5y=-15# #"Or"# #5y=3x+15# #y=3/5 x+3#

Abigail Allen · Answer

undefined To find the standard form equation of a line given two points, (0, 3) and (-5, 0), first, find the slope using the formula $m = \frac{y_2 - y_1}{x_2 - x_1}$, where $(x_1, y_1)$ and $(x_2, y_2)$ are the coordinates of the two points. Then, use the slope-intercept form $y = mx + b$ to find the y-intercept (b). Finally, rewrite the equation in standard form $Ax + By = C$, where A, B, and C are integers and A is positive.

Given points (0, 3) and (-5, 0):
$m = \frac{0 - 3}{-5 - 0} = \frac{-3}{-5} = \frac{3}{5}$

Using the point-slope form with the slope $m = \frac{3}{5}$ and the point (0, 3):
$y - 3 = \frac{3}{5}(x - 0)$
$y - 3 = \frac{3}{5}x$
$y = \frac{3}{5}x + 3$

To convert this to standard form, multiply every term by 5 to get rid of the fraction:
$5y = 3x + 15$

Rearrange the terms to have the x coefficient positive:
$3x - 5y = -15$

So, the standard form of the equation is $3x - 5y = -15$.