Question

graph the line with slope 3 passing through the point (-2, 5).

Explanation:

Step1: Recall point - slope form

The point - slope form of a line is $y - y_1=m(x - x_1)$, where $m$ is the slope and $(x_1,y_1)$ is a point on the line. Here, $m = 3$, $x_1=-2$, and $y_1 = 5$.
Substitute these values into the point - slope formula: $y - 5=3(x - (-2))$, which simplifies to $y - 5=3(x + 2)$.

Step2: Convert to slope - intercept form (optional for graphing)

Expand the right - hand side: $y - 5=3x+6$.
Add 5 to both sides: $y=3x + 6+5$, so $y=3x+11$.
To graph the line:

• Start with the point $(-2,5)$. Since the slope $m = 3=\frac{3}{1}$, from the point $(-2,5)$, we can move 1 unit to the right (increase $x$ by 1) and 3 units up (increase $y$ by 3) to get the next point $( - 2+1,5 + 3)=( - 1,8)$.
• We can also move 1 unit to the left (decrease $x$ by 1) and 3 units down (decrease $y$ by 3) from the point $(-2,5)$ to get the point $(-2 - 1,5-3)=(-3,2)$.
• Then draw a straight line passing through these points (and others obtained by using the slope) on the coordinate plane.

(Note: Since the problem is about graphing, the key steps are identifying the starting point and using the slope to find other points on the line.)

Answer:

To graph the line, plot the point \((-2,5)\), then use the slope \(3=\frac{3}{1}\) to find other points (e.g., \((-1,8)\) by moving 1 right and 3 up from \((-2,5)\), or \((-3,2)\) by moving 1 left and 3 down from \((-2,5)\)) and draw a straight line through them. The equation of the line is \(y = 3x+11\) (if needed for reference).