Line in a rectangular coordinate system
“write an equation that represents the line. use exact numbers” fits the algebraic task of expressing a straight line in a Cartesian plane as an exact linear equation. A specific line is required; the assumed line passes through the two plotted points \(A(-5,2)\) and \(B(4,-1)\).
Slope from two points
A non-vertical line has slope \(m\) given by \[ m = \frac{y_2 - y_1}{x_2 - x_1}. \]
Using \(A(-5,2)\) and \(B(4,-1)\), \[ m = \frac{-1 - 2}{4 - (-5)} = \frac{-3}{9} = -\frac{1}{3}. \]
Point-slope form and slope-intercept form
Point-slope form with point \(A(-5,2)\) is \[ y - 2 = -\frac{1}{3}(x + 5). \]
Expanding to slope-intercept form, \[ y - 2 = -\frac{1}{3}x - \frac{5}{3} \quad \Rightarrow \quad y = -\frac{1}{3}x + \frac{1}{3}. \]
An exact equation representing the line is \(y = -\dfrac{1}{3}x + \dfrac{1}{3}\).
Standard form with integer coefficients
Clearing denominators by multiplying by \(3\) gives \(3y = -x + 1\), so the equivalent standard form is \[ x + 3y = 1. \]
Consistency properties
| Point | Substitution into \(y = -\dfrac{1}{3}x + \dfrac{1}{3}\) | Result |
|---|---|---|
| \(A(-5,2)\) | \(y = -\dfrac{1}{3}(-5) + \dfrac{1}{3} = \dfrac{5}{3} + \dfrac{1}{3} = 2\) | On the line |
| \(B(4,-1)\) | \(y = -\dfrac{1}{3}(4) + \dfrac{1}{3} = -\dfrac{4}{3} + \dfrac{1}{3} = -1\) | On the line |
Intercepts follow from the same exact equation: the x-intercept satisfies \(0 = -\dfrac{1}{3}x + \dfrac{1}{3}\), giving \(x = 1\), and the y-intercept is \(\left(0,\dfrac{1}{3}\right)\).
Graph and geometric meaning of the slope
Common pitfalls
The sign in \(y_2 - y_1\) and \(x_2 - x_1\) controls the sign of the slope; swapping the order in only one difference changes \(m\) incorrectly. Exact numbers require fractions to remain unreduced decimals; \( -\dfrac{1}{3}\) and \(\dfrac{1}{3}\) preserve the exact line representation.