Question

Graph a line with a slope of —3 that contains the poigt (4, -2).y65432-7 -6 -5 -4 -3 -2→ 22 3 4 5 6 71-2+-3+-4+-5-6

Answer 1

Answer:

Points on the line are;

$\begin{gathered} (0,10) \\ (2,4) \end{gathered}$

Graphing the line we have;

Explanation:

Given a line of slope -3 and passes through the point (4,-2).

$\begin{gathered} m=-3 \\ (x_1,y_1)=(4,-2) \end{gathered}$

Let us derive the equation of the line and graph the line;

Recall that;

$y-y_1=m(x-x_1)$

substituting the given values;

$\begin{gathered} y-(-2)=-3(x-4) \\ y+2=-3x+12 \\ y=-3x+12-2 \\ y=-3x+10 \end{gathered}$

The equation of the line is;

$y=-3x+10$

Let us now derive the points on the line;

At x= 0;

$\begin{gathered} y=-3x+10 \\ y=0+10 \\ y=10 \\ (0,10)_{} \end{gathered}$

At x = 2;

$\begin{gathered} y=-3x+10 \\ y=-3(2)+10 \\ y=-6+10 \\ y=4 \\ (2,4) \end{gathered}$

Points on the line are;

$\begin{gathered} (0,10) \\ (2,4) \end{gathered}$

Graphing the line we have;