Final answer:
The line perpendicular to 3x + 5y = 1 and passing through (2,7) can be written in slope-intercept form as y = (5/3)x + (11/3) and in standard form as -5x + 3y = 11.
Step-by-step explanation:
To write an equation for a line that is perpendicular to another line and passes through a given point, you first need to find the slope of the original line. For the equation 3x + 5y = 1, we can rewrite it in slope-intercept form, which is y = mx + b, where m is the slope and b is the y-intercept. Rewriting it we get 5y = -3x + 1, which simplifies to y = (-3/5)x + (1/5). So, the slope of this line is -3/5. A line that is perpendicular to this line would have a slope that is the negative reciprocal of -3/5, which is 5/3.
Using the point (2,7) and the slope of 5/3, we can use the point-slope form of the equation (y - y1) = m(x - x1) to find our perpendicular line: (y - 7) = (5/3)(x - 2). Expanding this, we have y - 7 = (5/3)x - (10/3). Now we can add 7 to both sides to write it in slope-intercept form, y = (5/3)x + (11/3).
To write the equation in standard form (Ax + By = C), we need to get rid of the fractions and have A, B, and C as integers. Multiplying the entire equation by 3 gives us 3y = 5x + 11. To finish putting this into standard form, we can bring the x term to the left side: -5x + 3y = 11. This is the standard form of the equation.