Write an equation of the line containing the specified point and perpendicular to the indicated line. (6,0), 5x+4y=1
first line: 5x + 4y = 1
perpendicular line must be : 4x - 5y = c
but (6,0) lies on it, so ...
4(6) + 0 = c = 24
4x - 5y = 24
thank you
To find an equation of the line containing a specified point and perpendicular to a given line, follow these steps:
1. Determine the slope of the given line by rearranging the equation into slope-intercept form (y = mx + b), where m represents the slope.
Given line: 5x + 4y = 1
Rearrange the equation: 4y = -5x + 1
Divide by 4: y = -5/4*x + 1/4
The slope of the given line is -5/4.
2. The slope of a line perpendicular to the given line will be the negative reciprocal of the slope of the given line. To find the negative reciprocal, take the reciprocal (flip the fraction) and negate it (multiply by -1).
Negative reciprocal of -5/4 is 4/5.
3. Use the point-slope form of a linear equation (y - y₁ = m(x - x₁)), where (x₁, y₁) represents the given point and m is the slope of the perpendicular line.
Given point: (6, 0)
Slope of the perpendicular line: 4/5
Applying the point-slope form:
y - 0 = 4/5(x - 6)
4. Simplify the equation by distributing the 4/5:
y = 4/5x - 24/5
Therefore, the equation of the line containing the point (6,0) and perpendicular to the line 5x + 4y = 1 is y = 4/5x - 24/5.