To find the equations of straight lines passing through (1,1) and at a distance of 3 units from (-2,3), we can use the concept of perpendicular distance from a point to a line.



The slope of the line passing through two points (x1, y1) and (x2, y2) can be calculated using the formula:

Slope (m) = (y2 - y1) / (x2 - x1)

Using the given points (1,1) and (-2,3), the slope of the line passing through them is:

m = (3 - 1) / (-2 - 1) = 2 / -3 = -2/3



Since the line we are looking for is at a distance of 3 units from (-2,3), it is perpendicular to the line passing through (1,1) and (-2,3).

The slope of a line perpendicular to a given line is the negative reciprocal of its slope. Therefore, the slope of the perpendicular line is:

m_perpendicular = -1 / m = -1 / (-2/3) = 3/2



We now have the slope of the perpendicular line and a point it passes through (1,1). Using the point-slope form of the equation of a line:

y - y1 = m_perpendicular(x - x1)

Substituting the values of (x1, y1) = (1,1) and m_perpendicular = 3/2:

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

Simplifying the equation:

2y - 2 = 3x - 3
3x - 2y + 1 = 0

Therefore, the equation of the perpendicular line passing through (1,1) is 3x - 2y + 1 = 0.



Since the line we are looking for is at a distance of 3 units from (-2,3), it can be either parallel or perpendicular to the line passing through (1,1) and (-2,3).

To find the equation of the parallel line, we use the same slope as the line passing through (1,1) and (-2,3), which is -2/3. Using the point-slope form with the point (1,1):

y - y1 = m_parallel(x - x1)

Substituting the values of (x1, y1) = (1,1) and m_parallel = -2/