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To determine the equation of a line perpendicular to [tex]\( y + 5 = \frac{2}{3}(x - 11) \)[/tex] that passes through the point [tex]\( (5, -6) \)[/tex], we can follow a detailed, step-by-step process:
1. Rewrite the Given Line in Slope-Intercept Form:
- We need to express the given line in the form [tex]\( y = mx + b \)[/tex], where [tex]\( m \)[/tex] is the slope and [tex]\( b \)[/tex] is the y-intercept.
- Start with the given equation: [tex]\( y + 5 = \frac{2}{3}(x - 11) \)[/tex].
- Distribute [tex]\( \frac{2}{3} \)[/tex]: [tex]\( y + 5 = \frac{2}{3}x - \frac{2}{3} \cdot 11 \)[/tex].
- Simplify: [tex]\( y + 5 = \frac{2}{3}x - \frac{22}{3} \)[/tex].
- Isolate [tex]\( y \)[/tex]: [tex]\( y = \frac{2}{3}x - \frac{22}{3} - 5 \)[/tex].
- Convert 5 to a fraction: [tex]\( y = \frac{2}{3}x - \frac{22}{3} - \frac{15}{3} \)[/tex].
- Combine like terms: [tex]\( y = \frac{2}{3}x - \frac{37}{3} \)[/tex].
Thus, the slope [tex]\( m \)[/tex] of the given line is [tex]\( \frac{2}{3} \)[/tex].
2. Determine the Perpendicular Slope:
- The slope of a line perpendicular to another is the negative reciprocal of the original slope.
- Original slope [tex]\( m = \frac{2}{3} \)[/tex].
- Perpendicular slope [tex]\( m_{\perp} = -\frac{1}{m} = -\frac{1}{\frac{2}{3}} = -\frac{3}{2} \)[/tex].
3. Using Point-Slope Form to Find the Equation:
- The point-slope form of a line is [tex]\( y - y_1 = m(x - x_1) \)[/tex], where [tex]\( (x_1, y_1) \)[/tex] is a given point on the line and [tex]\( m \)[/tex] is the slope.
- Given point [tex]\( (5, -6) \)[/tex], and perpendicular slope [tex]\( -\frac{3}{2} \)[/tex].
- Substitute into the point-slope form: [tex]\( y - (-6) = -\frac{3}{2}(x - 5) \)[/tex].
- Simplify: [tex]\( y + 6 = -\frac{3}{2}(x - 5) \)[/tex].
- Distribute the slope: [tex]\( y + 6 = -\frac{3}{2}x + \frac{15}{2} \)[/tex].
- Isolate [tex]\( y \)[/tex]: [tex]\( y = -\frac{3}{2}x + \frac{15}{2} - 6 \)[/tex].
- Convert 6 to a fraction: [tex]\( y = -\frac{3}{2}x + \frac{15}{2} - \frac{12}{2} \)[/tex].
- Combine like terms: [tex]\( y = -\frac{3}{2}x + \frac{3}{2} \)[/tex].
Thus, the equation of the line in slope-intercept form ([tex]\( y = mx + b \)[/tex]) that is perpendicular to [tex]\( y + 5 = \frac{2}{3}(x - 11) \)[/tex] and passes through the point [tex]\( (5, -6) \)[/tex] is:
[tex]\[ y = -\frac{3}{2}x + \frac{3}{2} \][/tex]
This matches the option [tex]\( y = -\frac{3}{2}x + \frac{3}{2} \)[/tex].
1. Rewrite the Given Line in Slope-Intercept Form:
- We need to express the given line in the form [tex]\( y = mx + b \)[/tex], where [tex]\( m \)[/tex] is the slope and [tex]\( b \)[/tex] is the y-intercept.
- Start with the given equation: [tex]\( y + 5 = \frac{2}{3}(x - 11) \)[/tex].
- Distribute [tex]\( \frac{2}{3} \)[/tex]: [tex]\( y + 5 = \frac{2}{3}x - \frac{2}{3} \cdot 11 \)[/tex].
- Simplify: [tex]\( y + 5 = \frac{2}{3}x - \frac{22}{3} \)[/tex].
- Isolate [tex]\( y \)[/tex]: [tex]\( y = \frac{2}{3}x - \frac{22}{3} - 5 \)[/tex].
- Convert 5 to a fraction: [tex]\( y = \frac{2}{3}x - \frac{22}{3} - \frac{15}{3} \)[/tex].
- Combine like terms: [tex]\( y = \frac{2}{3}x - \frac{37}{3} \)[/tex].
Thus, the slope [tex]\( m \)[/tex] of the given line is [tex]\( \frac{2}{3} \)[/tex].
2. Determine the Perpendicular Slope:
- The slope of a line perpendicular to another is the negative reciprocal of the original slope.
- Original slope [tex]\( m = \frac{2}{3} \)[/tex].
- Perpendicular slope [tex]\( m_{\perp} = -\frac{1}{m} = -\frac{1}{\frac{2}{3}} = -\frac{3}{2} \)[/tex].
3. Using Point-Slope Form to Find the Equation:
- The point-slope form of a line is [tex]\( y - y_1 = m(x - x_1) \)[/tex], where [tex]\( (x_1, y_1) \)[/tex] is a given point on the line and [tex]\( m \)[/tex] is the slope.
- Given point [tex]\( (5, -6) \)[/tex], and perpendicular slope [tex]\( -\frac{3}{2} \)[/tex].
- Substitute into the point-slope form: [tex]\( y - (-6) = -\frac{3}{2}(x - 5) \)[/tex].
- Simplify: [tex]\( y + 6 = -\frac{3}{2}(x - 5) \)[/tex].
- Distribute the slope: [tex]\( y + 6 = -\frac{3}{2}x + \frac{15}{2} \)[/tex].
- Isolate [tex]\( y \)[/tex]: [tex]\( y = -\frac{3}{2}x + \frac{15}{2} - 6 \)[/tex].
- Convert 6 to a fraction: [tex]\( y = -\frac{3}{2}x + \frac{15}{2} - \frac{12}{2} \)[/tex].
- Combine like terms: [tex]\( y = -\frac{3}{2}x + \frac{3}{2} \)[/tex].
Thus, the equation of the line in slope-intercept form ([tex]\( y = mx + b \)[/tex]) that is perpendicular to [tex]\( y + 5 = \frac{2}{3}(x - 11) \)[/tex] and passes through the point [tex]\( (5, -6) \)[/tex] is:
[tex]\[ y = -\frac{3}{2}x + \frac{3}{2} \][/tex]
This matches the option [tex]\( y = -\frac{3}{2}x + \frac{3}{2} \)[/tex].
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