A line l passes through points (-2,3) and (-1,6) and perpendicular to a line p at (-1,6) Given that another line q is parallel to line one and passes through(1,2) find the x and y intersects?

First, let's find the slope of line p. The slope of a line perpendicular to line l passing through (-1,6) will be the negative reciprocal of the slope of line l.

The slope of line l passing through (-2,3) and (-1,6) is:
m = (6-3)/(-1-(-2)) = 3/1 = 3

The slope of line p will be the negative reciprocal of 3, which is -1/3.

Now, we can find the equation of line p passing through (-1,6) with slope -1/3.

Using the point-slope form of a linear equation:
y - y1 = m(x - x1)
y - 6 = (-1/3)(x - (-1))
y - 6 = (-1/3)(x + 1)
y - 6 = (-1/3)x - 1/3
y = (-1/3)x + 17/3

Multiplying through by 3 to clear the fraction:
3y = -x + 17

Rearranging to the form 9x + by = c:
9x + 3y = 17

Therefore, the equation of line p in the form 9x + by = c is 9x + 3y = 17.

Step 1: Find the slope of line l

Slope of line l = (y2 - y1) / (x2 - x1)
= (6 - 3) / (-1 - (-2))
= 3 / 1
= 3

Step 2: Since l is perpendicular to p, the slope of p is the negative reciprocal of 3.

Slope of p = -1/3

Step 3: Use the point-slope form to write the equation of p

Point-slope form: y - y1 = m(x - x1)
Where m is the slope and (x1, y1) is a point on the line.

Using the point (-1,6) and the slope -1/3, we get:
y - 6 = (-1/3)(x + 1)

Step 4: Convert the equation to the form 9x by = c

Multiply both sides by 9:
9(y - 6) = 9(-1/3)(x + 1)
9y - 54 = -3(x + 1)
9y - 54 = -3x - 3
9y = -3x - 51
9x + 9y = -51
Therefore, the equation of p is 9x + 9y = -51.