How do you find a standard form equation for the line with (-5, -16) with a slope of 2?

2 Answers
Jan 20, 2017

#color(red)(2)x - color(blue)(1)y = color(green)(6)#

Explanation:

First, we use the point-slope formula to write an equation for the line:

The point-slope formula states: #(y - color(red)(y_1)) = color(blue)(m)(x - color(red)(x_1))#

Where #color(blue)(m)# is the slope and #color(red)(((x_1, y_1)))# is a point the line passes through.

#(y - color(red)(-16)) = color(blue)(2)(x - color(red)(-5))#

#(y + color(red)(16)) = color(blue)(2)(x + color(red)(5))#

Now we can transform to the standard form.

The standard form of a linear equation is:

#color(red)(A)x + color(blue)(B)y = color(green)(C)#

where, if at all possible, #color(red)(A)#, #color(blue)(B)#, and #color(green)(C)#are integers, and A is non-negative, and, A, B, and C have no common factors other than 1

#y + color(red)(16) = (color(blue)(2) xx x) + (color(blue)(2) xx color(red)(5))#

#y + color(red)(16) = 2x + 10#

#y + color(red)(16) - 16 - color(blue)(2x) = 2x + 10 - 16 - color(blue)(2x)#

# -color(blue)(2x) + y + color(red)(16) - 16 = 2x - color(blue)(2x) + 10 - 16#

# -color(blue)(2x) + y + 0 = 0 + 10 - 16#

#-color(blue)(2x) + y = -6#

#-1(-color(blue)(2x) + y) = -1 xx -6#

#2x - y = 6#

#color(red)(2)x - color(blue)(1)y = color(green)(6)#

Jan 20, 2017

#2x-y-6=0#

Explanation:

The #color(blue)"standard form equation "# of a line is.

#color(red)(bar(ul(|color(white)(2/2)color(black)(ax+by+c=0)color(white)(2/2)|)))#

Before this, we can write the equation in #color(blue)"point-slope form"#

#color(red)(bar(ul(|color(white)(2/2)color(black)(y-y_1=m(x-x_1))color(white)(2/2)|)))#
where m represents the slope and # (x_1,y_1)" a point on the line"#

here #m=2" and " (x_1,y_1)=(-5,-16)#

#rArry-(-16)=2(x-(-5))#

#rArry+16=2(x+5)#

#rArry+16=2x+10larrcolor(red)" in point-slope form"#

Rearranging in standard form.

#rArr2x-y-6=0larrcolor(red)" in standard form"#