line containing poir

Elementary Geometry For College Students, 7e
7th Edition
ISBN:9781337614085
Author:Alexander, Daniel C.; Koeberlein, Geralyn M.
Publisher:Alexander, Daniel C.; Koeberlein, Geralyn M.
ChapterP: Preliminary Concepts
SectionP.CT: Test
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Write an equation for the lines described below in all 3 forms (point-slope, slope intercept, and standard form.
### Problem Description

**10. A Line Containing Points (3, -4) and (-3, 4)**

Given two points on a Cartesian coordinate plane, (3, -4) and (-3, 4), find the equation of the line that passes through these points.

### Explanation
First, we need to determine the slope (\(m\)) of the line that contains the given points:

\[ m = \frac{{y_2 - y_1}}{{x_2 - x_1}} \]

Let's denote the points as follows:
- \( (x_1, y_1) = (3, -4) \)
- \( (x_2, y_2) = (-3, 4) \)

Now, plugging in the coordinates into the slope formula:

\[ m = \frac{{4 - (-4)}}{{-3 - 3}} = \frac{{4 + 4}}{{-3 - 3}} = \frac{8}{-6} = -\frac{4}{3} \]

The slope of the line is \( -\frac{4}{3} \).

Next, we use the point-slope form of the equation of a line, which is given by:

\[ y - y_1 = m(x - x_1) \]

Let's use the point (3, -4):

\[ y - (-4) = -\frac{4}{3}(x - 3) \]

Simplifying, we find:

\[ y + 4 = -\frac{4}{3}(x - 3) \]
\[ y + 4 = -\frac{4}{3}x + 4 \]

Subtract 4 from both sides to solve for \(y\):

\[ y = -\frac{4}{3}x \]

Thus, the equation of the line passing through the points (3, -4) and (-3, 4) is:

\[ y = -\frac{4}{3}x \]
Transcribed Image Text:### Problem Description **10. A Line Containing Points (3, -4) and (-3, 4)** Given two points on a Cartesian coordinate plane, (3, -4) and (-3, 4), find the equation of the line that passes through these points. ### Explanation First, we need to determine the slope (\(m\)) of the line that contains the given points: \[ m = \frac{{y_2 - y_1}}{{x_2 - x_1}} \] Let's denote the points as follows: - \( (x_1, y_1) = (3, -4) \) - \( (x_2, y_2) = (-3, 4) \) Now, plugging in the coordinates into the slope formula: \[ m = \frac{{4 - (-4)}}{{-3 - 3}} = \frac{{4 + 4}}{{-3 - 3}} = \frac{8}{-6} = -\frac{4}{3} \] The slope of the line is \( -\frac{4}{3} \). Next, we use the point-slope form of the equation of a line, which is given by: \[ y - y_1 = m(x - x_1) \] Let's use the point (3, -4): \[ y - (-4) = -\frac{4}{3}(x - 3) \] Simplifying, we find: \[ y + 4 = -\frac{4}{3}(x - 3) \] \[ y + 4 = -\frac{4}{3}x + 4 \] Subtract 4 from both sides to solve for \(y\): \[ y = -\frac{4}{3}x \] Thus, the equation of the line passing through the points (3, -4) and (-3, 4) is: \[ y = -\frac{4}{3}x \]
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