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Find the equation of the given line. Slope is \( \frac{2}{3} ;(6,3) \) on line

Ask by Hampton Montgomery. in the United States
Jan 24,2025

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The equation of the line is \( y = \frac{2}{3}x - 1 \).

Solución

To find the equation of a line with a given slope and a point on the line, you can use the **point-slope form** of a linear equation: \[ y - y_1 = m(x - x_1) \] Where: - \( m \) is the slope. - \( (x_1, y_1) \) is the given point on the line. **Given:** - Slope (\( m \)) = \( \frac{2}{3} \) - Point (\( x_1, y_1 \)) = \( (6, 3) \) **Step 1: Plug the values into the point-slope formula.** \[ y - 3 = \frac{2}{3}(x - 6) \] **Step 2: Simplify the equation to get it into slope-intercept form (\( y = mx + b \)).** \[ y - 3 = \frac{2}{3}x - \frac{2}{3} \times 6 \] \[ y - 3 = \frac{2}{3}x - 4 \] \[ y = \frac{2}{3}x - 4 + 3 \] \[ y = \frac{2}{3}x - 1 \] **Final Equation:** \[ y = \frac{2}{3}x - 1 \] **Alternatively, in Standard Form (\( Ax + By = C \)):** Multiply both sides by 3 to eliminate the fraction: \[ 3y = 2x - 3 \] \[ 2x - 3y = 3 \] So, the equation of the line is either: - **Slope-Intercept Form:** \( y = \frac{2}{3}x - 1 \) - **Standard Form:** \( 2x - 3y = 3 \)

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To find the equation of the line, we can use the point-slope form of the equation \( y - y_1 = m(x - x_1) \), where \( m \) is the slope and \((x_1, y_1)\) is a point on the line. Given the slope \( m = \frac{2}{3} \) and the point \((6, 3)\): \[ y - 3 = \frac{2}{3}(x - 6) \] Now, let's simplify that: \[ y - 3 = \frac{2}{3}x - 4 \] \[ y = \frac{2}{3}x - 1 \] So, the equation of the line is \( y = \frac{2}{3}x - 1 \). If you have ever wondered how slopes work in the real world, think of them as the steepness of a hill. A slope of \( \frac{2}{3} \) means that for every 3 units you move horizontally, you move up 2 units vertically. This is like walking up a gentle hill, where the incline is not too steep, making it manageable to climb! When using point-slope form, a common mistake is forgetting to include the point coordinates in the equation. Be careful to substitute the correct \( x_1 \) and \( y_1 \) values from your point, as it ensures that your line passes through the specified point, maintaining accuracy in your calculations.

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