Question

1. for a line whose slope is -3 and which passes through the point (5, -2): (a) write the equation of this line in point - slope form, y - y₁ = m(x - x₁). (b) write the equation of this line in slope - intercept form, y = mx + b. 5. a line passes through the points (5, - 4) and (20, 8). (a) write the equation of this line in point - slope form, y - y₁ = m(x - x₁). (b) write the equation of this line in slope - intercept form, y = mx + b.

Explanation:

Step1: Recall point - slope formula

The point - slope form is $y - y_1=m(x - x_1)$, where $m$ is the slope and $(x_1,y_1)$ is a point on the line.

Step2: For problem 4(a)

Given $m=-3$, $x_1 = 5$ and $y_1=-2$. Substitute into the point - slope formula:
$y-(-2)=-3(x - 5)$ which simplifies to $y + 2=-3(x - 5)$.

Step3: For problem 4(b)

Start with $y + 2=-3(x - 5)$. Expand the right - hand side: $y+2=-3x + 15$. Then solve for $y$:
$y=-3x+15 - 2$, so $y=-3x + 13$.

Step4: For problem 5(a)

First, find the slope $m=\frac{y_2-y_1}{x_2-x_1}$. Here, $x_1 = 5,y_1=-4,x_2 = 20,y_2 = 8$.
$m=\frac{8-(-4)}{20 - 5}=\frac{8 + 4}{15}=\frac{12}{15}=\frac{4}{5}$.
Using the point $(x_1,y_1)=(5,-4)$ and the point - slope formula, we get $y-(-4)=\frac{4}{5}(x - 5)$, which simplifies to $y + 4=\frac{4}{5}(x - 5)$.

Step5: For problem 5(b)

Start with $y + 4=\frac{4}{5}(x - 5)$. Expand the right - hand side: $y+4=\frac{4}{5}x-4$. Then solve for $y$:
$y=\frac{4}{5}x-4 - 4$, so $y=\frac{4}{5}x-8$.

Answer:

4(a). $y + 2=-3(x - 5)$
4(b). $y=-3x + 13$
5(a). $y + 4=\frac{4}{5}(x - 5)$
5(b). $y=\frac{4}{5}x-8$