How do you write the equation in standard form given m = -1/2 and (4,6)?

Question

Hazel Anglin · Accepted Answer

#x+2y-16=0#

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

#color(red)(bar(ul(|color(white)(2/2)color(black)(ax+by+c=0)color(white)(2/2)|)))# where a is a positive integer and b, c are integers.

Begin by expressing 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)" is a point on the line"#

#"here "m=-1/2" and " (x_1,y_1)=(4,6)#

Substituting these values into the equation.

#y-6=-1/2(x-4)larrcolor(red)"point-slope form"#

distributing, simplifying and rearranging.

#y-6=-1/2x+2#

#rArr1/2x+y-8=0#

coefficient of x, that is a must be an integer.

multiplying through by 2

#rArrx+2y-16=0larrcolor(red)" in standard form"#

Jameson Allen · Answer

undefined To write the equation in standard form given $ m = -\frac{1}{2} $ and the point $ (4,6) $, use the point-slope form: $ y - y_1 = m(x - x_1) $. Substitute the given values into the equation and then manipulate it into standard form $ Ax + By = C $. The resulting equation is: $ y - 6 = -\frac{1}{2}(x - 4) $. Expanding and rearranging terms gives $ 2y - 12 = -x + 4 $, and further rearranging yields $ x + 2y = 16 $, which is the equation in standard form.