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Assume that \(a\), \(b\), and \(c\) are constant real numbers: *parameters.*

The solution to the equation \(a\cdot x+b\cdot y+c=0\) can be drawn in the plane. They are the points \(\left[x,y\right]\) satisfying \(a\cdot x+b\cdot y+c=0\). If \(a\ne0\) or \(b\ne0\), then these points form a straight line, or simply just a line.

- If \(b\ne0\), then the equation can be written as \(y=-\frac{a}{b}\cdot{x}-\frac{c}{b}\). For, these are the solutions if we consider \(x\) as a parameter and \(y\) as unknown. This indicates that for every value of \(x\) there is a point \(\left[x,y\right]\) with \(y\) equal to \(-\frac{a}{b}\cdot{x}-\frac{c}{b}-\frac{a}{b}\cdot{x}-\frac{c}{b}\).
- If \(a\ne0\), the line is oblique (by oblique we mean neither horizontal nor vertical).
- If \(a = 0\), then the value of \(y\) is constant, equal to \(-\frac{c}{b}\). In this case the line is horizontal.

- In the exceptional case \(b = 0\) the equation looks like \(a \cdot{x}+c = 0\).
- If \(a\ne0\), then the line is vertical.
- If \(a = 0\) and
- \(c\ne0\), then there are no solutions
- \(c = 0\), then each pair of values of \(\left[x,y\right]\) is a solution.

A straight line can be described in different ways.

- The solutions \(\left[x,y\right]\) to an equation \(a\cdot x+b\cdot y+c=0\) with unknowns \(x\) and \(y\). Here \(a\), \(b\) and \(c\) are real numbers such that \(a\) and \(b\) are not equal to zero.
- The line through two given points in the plane: if \(P = \left[x,y\right]\) and \(Q = \left[s,t\right]\) are points in the plane, then the line through \(P\) and \(Q\) has equation \(a\cdot x+b\cdot y+c=0\) with \(a=q-t\), \(b = s – p\) and \(c = t \cdot {p} – q \cdot{s}\).
- The line through a given point, the base point, and a direction, indicated by the number \(-\frac{a}{b}\), where \(a\) and \(b\) are as in the equation given above; this number is called the slope of the line.
- The line with function representation \(y = p\cdot x+q\) if \(b\ne0\) and \(x = r\) otherwise; here we have \(p = -\frac{a}{b}\) (the slope), \(q = -\frac{c}{b}\) (the intercept), which is the value of \(y\) for \(x = 0\) and \(r = -\frac{c}{a}\) in terms of the above \(a\), \(b\) and \(c\). This can be seen as a special case of the previous description, with base point \(\left[0,q\right]\). In the case where \(b\ne0\), the variable \(y\) is a function of \(x\), in the other case, \(x\) is a constant function of \(y\).

**Let’s practice!**

The line segment between the two points \(\left[2,\frac{77}{12}\right]\) and \(\left[6,\frac{63}{4}\right]\) is drawn in the figure below.

Give the function rule for this line in the form \(y = a\cdot{x} + b\).

The line is described by the function rule \(y = a\cdot{x} + b\) where \(a\) is the slope, given as the quotient of the difference of the \(y\)-values with the difference of the \(x\)-values of two points on the line. Hence \(a=\frac{{{63}\over{4}}-{{77}\over{12}}}{6-2}={{7}\over{3}}\). The value of \(b\) follows from \(b = y – a\cdot{x}\), where \(\left[x,y\right]\) is a random point on the line. Hence \(b={{77}\over{12}}-{{7}\over{3}}\cdot{2}={{7}\over{4}}\).

**Answer:**

\(y={{7}\over{3}}\cdot{x}+{{7}\over{4}}\)