Answer :
We are given two points: [tex]$(-2, 3)$[/tex] and [tex]$(-1, 6)$[/tex]. The goal is to find:
1. The gradient (slope) of the line [tex]$L_1$[/tex].
2. The equation of the line [tex]$L_1$[/tex] in slope-intercept form.
We start with the formula for the gradient of a line that passes through the points [tex]$(x_1, y_1)$[/tex] and [tex]$(x_2, y_2)$[/tex]:
[tex]$$
m = \frac{y_2 - y_1}{x_2 - x_1}.
$$[/tex]
Here, let [tex]$(x_1, y_1)=(-2, 3)$[/tex] and [tex]$(x_2, y_2)=(-1, 6)$[/tex]. Then:
[tex]$$
m = \frac{6 - 3}{-1 - (-2)} = \frac{3}{1} = 3.
$$[/tex]
Thus, the gradient of [tex]$L_1$[/tex] is [tex]$3$[/tex].
Next, we use the slope-intercept form of a line:
[tex]$$
y = mx + c,
$$[/tex]
where [tex]$m$[/tex] is the gradient and [tex]$c$[/tex] is the y-intercept. We already found [tex]$m=3$[/tex]. To determine [tex]$c$[/tex], substitute one of the points into the equation. Using the point [tex]$(-2, 3)$[/tex]:
[tex]$$
3 = 3(-2) + c.
$$[/tex]
Solve for [tex]$c$[/tex]:
[tex]$$
3 = -6 + c \quad \Longrightarrow \quad c = 3 + 6 = 9.
$$[/tex]
Therefore, the y-intercept is [tex]$9$[/tex], and the equation of the line is:
[tex]$$
y = 3x + 9.
$$[/tex]
In summary:
- The gradient of [tex]$L_1$[/tex] is [tex]$3$[/tex].
- The equation of [tex]$L_1$[/tex] is [tex]$y = 3x + 9$[/tex].
1. The gradient (slope) of the line [tex]$L_1$[/tex].
2. The equation of the line [tex]$L_1$[/tex] in slope-intercept form.
We start with the formula for the gradient of a line that passes through the points [tex]$(x_1, y_1)$[/tex] and [tex]$(x_2, y_2)$[/tex]:
[tex]$$
m = \frac{y_2 - y_1}{x_2 - x_1}.
$$[/tex]
Here, let [tex]$(x_1, y_1)=(-2, 3)$[/tex] and [tex]$(x_2, y_2)=(-1, 6)$[/tex]. Then:
[tex]$$
m = \frac{6 - 3}{-1 - (-2)} = \frac{3}{1} = 3.
$$[/tex]
Thus, the gradient of [tex]$L_1$[/tex] is [tex]$3$[/tex].
Next, we use the slope-intercept form of a line:
[tex]$$
y = mx + c,
$$[/tex]
where [tex]$m$[/tex] is the gradient and [tex]$c$[/tex] is the y-intercept. We already found [tex]$m=3$[/tex]. To determine [tex]$c$[/tex], substitute one of the points into the equation. Using the point [tex]$(-2, 3)$[/tex]:
[tex]$$
3 = 3(-2) + c.
$$[/tex]
Solve for [tex]$c$[/tex]:
[tex]$$
3 = -6 + c \quad \Longrightarrow \quad c = 3 + 6 = 9.
$$[/tex]
Therefore, the y-intercept is [tex]$9$[/tex], and the equation of the line is:
[tex]$$
y = 3x + 9.
$$[/tex]
In summary:
- The gradient of [tex]$L_1$[/tex] is [tex]$3$[/tex].
- The equation of [tex]$L_1$[/tex] is [tex]$y = 3x + 9$[/tex].
