Question

What is an equation of the line that passes through
the points (4, -3) and (2, 0)?

Answer 1

y = -1.5x + 3

Explanation:

gradient = (-3 - 0) / (4 - 2) = -3/2 = -1.5.

equation on line through two points is given by

y - y1 = m (x - x1)

where y1 and x1 are the coordinates of the point and m is the gradient.

we can choose either point to use in the equation.

y - 0 = -1.5 (x - 2) = -1.5x + 3

y = -1.5x + 3

Answer 2

`(\stackrel{x_1}{4}~,~\stackrel{y_1}{-3})\qquad (\stackrel{x_2}{2}~,~\stackrel{y_2}{0}) \\\\\\ \stackrel{slope}{m}\implies \cfrac{\stackrel{\textit{\large rise}} {\stackrel{y_2}{0}-\stackrel{y1}{(-3)}}}{\underset{\textit{\large run}} {\underset{x_2}{2}-\underset{x_1}{4}}} \implies \cfrac{0 +3}{-2} \implies \cfrac{ 3 }{ -2 } \implies - \cfrac{3}{2}`

`\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-\stackrel{y_1}{(-3)}=\stackrel{m}{- \cfrac{3}{2}}(x-\stackrel{x_1}{4}) \implies y +3 = - \cfrac{3}{2} ( x -4) \\\\\\ y+3=- \cfrac{3}{2}x+6\implies {\Large \begin{array}{llll} y=- \cfrac{3}{2}x+3 \end{array}}`