Find the equation of the line passing through the point R ( 1,-5) and perpendicular to the line 3x-y+11=0?

To find the equation of the line passing through point R (1, -5) and perpendicular to the line 3x - y + 11 = 0, we first need to find the slope of the given line.

The given line can be rewritten in slope-intercept form (y = mx + b) by solving for y:
y = 3x + 11

The slope of the given line 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 (1, -5), we can use the point-slope form of the equation of a line to find the equation of the new line:
y - y1 = m(x - x1)

Substitute the values of the point R (1, -5) and the slope (-1/3) into the equation:
y - (-5) = (-1/3)(x - 1)
y + 5 = (-1/3)x + 1/3

Now, simplify the equation:
y = (-1/3)x + 1/3 - 5
y = (-1/3)x - 14/3

Therefore, the equation of the line passing through point R (1, -5) and perpendicular to the line 3x - y + 11 = 0 is y = (-1/3)x - 14/3.

Step 1: Find the slope of the given line.

The given line is 3x - y + 11 = 0. Rewrite it in slope-intercept form y = mx + c:

y = 3x - 11
y = 3x + (-11)

The slope (m) of the given line is 3.

Step 2: Find the slope of the perpendicular line.

The slope of a line perpendicular to another line is the negative reciprocal of the original slope.

Slope of perpendicular line = -1/3

Step 3: Use the point-slope form of a line.

The point-slope form of a line is:

y - y1 = m(x - x1)

where (x1, y1) is a point on the line and m is the slope.

Using the point R(1, -5) and the slope -1/3, the equation of the line passing through R and perpendicular to the given line is:

y - (-5) = -1/3(x - 1)
y + 5 = -1/3x + 1/3
y = -1/3x + 4/3