Final answer:
The equation of the line passing through the point (2,3) and perpendicular to the line 3x-5y=6 is y = (-5/3)x + 13/3.
Step-by-step explanation:
The question asks to find the equation of a line that passes through the point (2,3) and is perpendicular to the line with the equation 3x-5y=6. To find this, we first need to determine the slope of the original line by transforming its equation into slope-intercept form (y=mx+b), where m represents the slope and b represents the y-intercept.
Rearranging the given line equation we get 5y=3x-6, and then y=(3/5)x-(6/5). The slope of this line (m) is 3/5. For a line to be perpendicular to this one, its slope must be the negative reciprocal of 3/5, which is -5/3.
With the slope of the new line and the point (2,3), we can use the point-slope form of a line equation:
y - y1 = m(x - x1)
Plugging in the values for m (-5/3), x1 (2), and y1 (3) gives:
y - 3 = (-5/3)(x - 2)
This equation can be simplified and written in slope-intercept form:
y = (-5/3)x + 13/3
This is the equation of the line that passes through the point (2,3) and is perpendicular to the line 3x-5y=6.