How to write an equation in standard form for a line through (4,3); slope 1/2?

Question

Isaiah Anthony · Accepted Answer

#x-2y=-2# #"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"# #"obtain the equation in "color(blue)"slope-intercept form"# #•color(white)(x)y=mx+b# #"where m is the slope and b the y-intercept"# #"here "m=1/2# #y=1/2x+blarrcolor(blue)"is the partial equation"# #"to find b substitute "(4,3)" into the partial equation"# #3=2+brArrb=3-2=1# #y=1/2x+1larrcolor(red)"in slope-intercept form"# #"multiply through by 2"# #2y=x+2# #x-2y=-2larrcolor(red)"in standard form"#

Eva Adamson · Answer

undefined To write an equation in standard form for a line through the point (4,3) with a slope of 1/2, use the point-slope form of a linear equation: y - y1 = m(x - x1), where (x1, y1) is the given point and m is the slope. Substituting the given values, we get y - 3 = (1/2)(x - 4). Then, simplify the equation: y - 3 = (1/2)x - 2. Multiply both sides by 2 to eliminate the fraction: 2(y - 3) = x - 4. Distribute 2 on the left side: 2y - 6 = x - 4. Rearrange the terms to write the equation in standard form: x - 2y = 2. Therefore, the equation of the line in standard form is x - 2y = 2.