Final answer:
To find the equation of a line perpendicular to 3x+y=10 that goes through (6,-5), first determine the negative reciprocal of the original line's slope (-3), which is 1/3. Then use the point-slope form with this new slope and the given point to derive the equation y = 1/3x - 7.
Step-by-step explanation:
The student is asking for the equation of a line that is perpendicular to the given line 3x+y=10 and passes through the point (6,-5).
Firstly, we need to find the slope of the given line. Rearrange the equation to the slope-intercept form, y = mx + b, where m represents the slope. The equation becomes y = -3x + 10, so the slope of the given line is -3.
The slope of the line that is perpendicular to this line will be the negative reciprocal of -3, which is 1/3.
Using the point-slope form of the equation of a line, y - y1 = m(x - x1), where (x1, y1) is the point through which the line passes and m is the slope, we can plug in the slope 1/3 and the point (6,-5) to get the equation of the new line: y - (-5) = 1/3(x - 6), which simplifies to y + 5 = 1/3x - 2.
Finally, we put it into slope-intercept form to get the desired equation: y = 1/3x - 7.