Question

Write an equation for the line described. give the answer in standard form. through (3,1),m=-6

Answer 1

`\bf (\stackrel{x_1}{3}~,~\stackrel{y_1}{1})~\hspace{10em} slope = m\implies -6 \\\\\\ \begin{array}ll \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-1=-6(x-3) \\\\\\ y-1=-6x+18\implies y=-6x+19`

bearing in mind that

standard form for a linear equation means

• all coefficients must be integers, no fractions

• only the constant on the right-hand-side

• all variables on the left-hand-side, sorted

• "x" must not have a negative coefficient

then we'd end up with 6x + y = 19.

Answer 2

6x + y = 19

Explanation:

The point-slope formula for a straight line is

y – y₁ = m(x – x₁)

x₁ = 3; y₁ = 1; m = -6 Substitute the values

y – 1 = -6(x-3) Remove parentheses

y – 1 = -6x + 18 Add 1 to each side

y = -6x + 19 Add 6x to each side

6x + y = 19

The graph is a straight line with a y-intercept at y = 19 and

slope = -18/3 = -6.