Find the equation of the straight line that passes through (2,-3) and is perpendicular to y-3x-1=0?

To find the equation of the line that is perpendicular to y-3x-1=0 and passes through the point (2,-3), we first need to find the slope of the given line.

The given line can be rewritten in the form y = 3x + 1. This is in the form y = mx + b, where m is the slope of the line. In this case, the slope is 3.

Since the line we are looking for is perpendicular to the given line, the slope of the new line will be the negative reciprocal of the slope of the given line. Therefore, the slope of the new line is -1/3.

Now that we have the slope of the new line and a point it passes through (2,-3), we can use the point-slope form of the equation of a line to find the equation of the new line.

The point-slope form of the equation of a line is y - y1 = m(x - x1), where (x1, y1) is the point the line passes through and m is the slope of the line.

Plugging in the values we have, we get:

y - (-3) = -1/3(x - 2)
y + 3 = -1/3x + 2/3
y = -1/3x + 2/3 - 3
y = -1/3x - 7/3

Therefore, the equation of the straight line that passes through (2,-3) and is perpendicular to y-3x-1=0 is y = -1/3x - 7/3.

The given equation can be written in the slope-intercept form as

$$ y - 3x - 1 = 0$$

$$\Rightarrow y = 3x + 1$$

So, the slope of the given line is 3.

Since the required line is perpendicular to the given line, its slope will be the negative reciprocal of 3 which is -1/3.

Now, using the point-slope formula, the equation of the required line passing through (2,-3) with slope -1/3 is given by

$$y - (-3) = -\frac{1}{3}(x - 2)$$

$$y + 3 = -\frac{1}{3}x + \frac{2}{3}$$

$$y = -\frac{1}{3}x + \frac{2}{3} - 3$$

$$y = -\frac{1}{3}x - \frac{7}{3}$$

Thus, the equation of the required line is $$y = -\frac{1}{3}x - \frac{7}{3}$$