How does slope intercept form differ from standard form?

Question

Savannah Anderson · Accepted Answer

Slope intercept form:
#y = mx + c# where #m# is the slope and #c# the y-intercept Thus, this form only displays a line's gradient and intercept. Whereas, the standard from:
#ax + bx + c = 0# where #a, b # and #c# are just constants.
This form doesn't show you gradient nor intercept,
on the other hand its useful when applying some formulae such as, the distance measured perpendicularly between a line and a point: Distance = #|(ax + by + c)/sqrt(a^2 + b^2)|#

Isaiah Anthony · Answer

undefined The slope-intercept form of a linear equation is written as y = mx + b, where m represents the slope of the line, and b represents the y-intercept. In this form, it explicitly shows the slope and y-intercept of the line.

On the other hand, the standard form of a linear equation is written as Ax + By = C, where A, B, and C are constants, and A and B are not both zero. This form does not directly reveal the slope and y-intercept; instead, it represents the equation in a more generalized form.

In summary, while the slope-intercept form explicitly displays the slope and y-intercept of a line, the standard form provides a more general representation of a linear equation.

Eli Assouline · Answer

undefined Slope-intercept form of a linear equation is $y = mx + b$, where $m$ represents the slope of the line and $b$ represents the y-intercept. Standard form of a linear equation is $Ax + By = C$, where $A$, $B$, and $C$ are constants and $A$ is typically positive. The main difference is that slope-intercept form directly reveals the slope and y-intercept of the line, while standard form does not immediately provide this information. Additionally, in slope-intercept form, the equation is explicitly solved for $y$, while in standard form, it is not necessarily solved for any specific variable.