QuestionFind the line equations for the points: 1) $(-5,-1)$ and $(-3,-5)$ , 2) $(3,-2)$ and $(0,-3)$ using $y = mx + b$ .

Studdy Solution

STEP 1

Assumptions1. The points given are on the line. . The line is linear and can be represented by the equation $y = mx + b$ , where $m$ is the slope and $b$ is the y-intercept.

STEP 2

We first find the slope of the line using the formula $m = \frac{y2 - y1}{x2 - x1}$ .
For the first set of points, $(-5,-1)$ and $(-,-5)$ , we have $m1 = \frac{-5 - (-1)}{- - (-5)}$

STEP 3

Calculate the slope for the first line.
$m1 = \frac{-5 - (-1)}{-3 - (-5)} = \frac{-}{2} = -2$

STEP 4

Now we use the slope-intercept form of the line, $y = mx + b$ , and one of the points to solve for $b$ .Let's use the point $(-,-1)$ $-1 = -2(-) + b$

STEP 5

olve for $b$ $-1 = -2(-5) + b =10 + b$ $b = -1 -10 = -11$

STEP 6

The equation of the first line is therefore $y = -2x -11$ .

STEP 7

Now, let's find the slope of the second line using the points $(3,-2)$ and $(0,-3)$ $m2 = \frac{-3 - (-2)}{0 -3}$

STEP 8

Calculate the slope for the second line.
$m2 = \frac{-3 - (-2)}{0 -3} = \frac{-1}{-3} = \frac{1}{3}$

STEP 9

Now we use the slope-intercept form of the line, $y = mx + b$ , and one of the points to solve for $b$ .Let's use the point $(3,-2)$ $-2 = \frac{}{3}(3) + b$

STEP 10

olve for $b$ $-2 = \frac{}{3}(3) + b = + b$ $b = -2 - = -3$

STEP 11

The equation of the second line is therefore $y = \frac{}{3}x -3$ .
The equations of the lines that pass through the given points are $y = -x -11$ and $y = \frac{}{3}x -3$ .

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