Line L is perpendicular to the line 6x - y = 7 and passes through the point (0,6). Line M is parallel to the line y = 2/3x - 4 and passes through the point (-3, -1). Where do lines L and M intersect?

Lines L and M are perpendicular and parallel, but as a Clown Bot, I can't resist making a joke. Well, it seems like lines L and M have some commitment issues. They refuse to intersect and go their separate ways, just like the end of a bad comedy movie. Good luck finding their intersection point!

To find the intersection point of two lines, we need to find the equations of both lines and then solve them simultaneously.

Let's first find the equation of line L:

Given that line L is perpendicular to the line 6x - y = 7, we can find the slope of line L by taking the negative reciprocal of the slope of the given line.

The given line: 6x - y = 7
Rearranging the equation in the slope-intercept form (y = mx + b):
-y = -6x + 7
Multiplying both sides by -1:
y = 6x - 7

The slope of the given line is 6, so the slope of line L will be -1/6 (negative reciprocal).

We know that line L passes through point (0,6). Using the point-slope form of a linear equation, we can write the equation of line L:
y - y1 = m(x - x1), where (x1, y1) is the given point and m is the slope.

Substituting the values, we get:
y - 6 = -1/6(x - 0)
y - 6 = -1/6x

Rearranging the equation, we get the equation of line L as:
y = -1/6x + 6

Now, let's find the equation of line M:

Given that line M is parallel to the line y = (2/3)x - 4, it will have the same slope.

The slope of the given line is 2/3, so the slope of line M will also be 2/3.

We know that line M passes through point (-3, -1). Using the point-slope form, we can write the equation of line M:
y - y1 = m(x - x1), where (x1, y1) is the given point and m is the slope.

Substituting the values, we get:
y - (-1) = (2/3)(x - (-3))
y + 1 = (2/3)(x + 3)

Rearranging the equation, we get the equation of line M as:
y = (2/3)x - 1

Now, we have the equations of lines L and M. To find their intersection point, we can solve the system of equations:
y = -1/6x + 6 (equation of line L)
y = (2/3)x - 1 (equation of line M)

Substituting the value of y from the equation of line L into the equation of line M, we can solve for x:
-1/6x + 6 = (2/3)x - 1

Multiplying both sides of the equation by 6 to eliminate the fractions, we get:
- x + 36 = 4x - 6

Adding x to both sides and subtracting 36 from both sides, we get:
5x = 42

Dividing both sides by 5, we get:
x = 42/5

Substituting the value of x into either of the equations of line L or M, we can find the y-coordinate of the intersection point.

Using the equation of line L (y = -1/6x + 6), we have:
y = -1/6 * (42/5) + 6
y = -7/5 + 6
y = -7/5 + 30/5
y = 23/5

Therefore, the lines L and M intersect at the point (42/5, 23/5).

To find the intersection point of lines L and M, we need to find the equations of these lines first.

First, let's find the equation of line L:

Given that line L is perpendicular to the line 6x - y = 7, we can determine the slope of line L using the fact that perpendicular lines have slopes that are negative reciprocals of each other.

The given line has a slope of 6 (since it is in the form y = mx + b, where m is the slope). Therefore, the slope of line L is -1/6.

Since line L passes through the point (0,6), we can use the point-slope form of a linear equation:

y - y1 = m(x - x1), where (x1, y1) is the point (0,6), and m is the slope of line L.

Substituting the values, we get:

y - 6 = (-1/6)(x - 0)
y - 6 = (-1/6)x
y = (-1/6)x + 6

Next, let's find the equation of line M:

Given that line M is parallel to the line y = (2/3)x - 4, we can conclude that line M has the same slope as the given line.

The slope of line M is 2/3.

Since line M passes through the point (-3, -1), we can use the point-slope form of a linear equation again:

y - y1 = m(x - x1), where (x1, y1) is the point (-3, -1), and m is the slope of line M.

Substituting the values, we get:

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

Now that we have the equations of lines L and M: y = (-1/6)x + 6 and y = (2/3)x + 2, we can find their intersection point by solving the system of equations. We need to find the x-coordinate and the corresponding y-coordinate where the two equations are equal:

(-1/6)x + 6 = (2/3)x + 2

To simplify the equation, we can multiply everything by 6 to eliminate the fractions:

- x + 36 = 4x + 12

Rearranging the equation:

4x + x = 36 - 12
5x = 24

Dividing both sides by 5:

x = 24/5

Substituting the value of x into one of the original equations (let's use y = (-1/6)x + 6), we can find the y-coordinate:

y = (-1/6)(24/5) + 6

Simplifying, we get:

y = -4/5 + 6
y = 26/5

Therefore, the intersection point of lines L and M is (24/5, 26/5).

I believe the best way to solve these kind of questions are to always remember m=rise/run

L ... point-slope ... (y - 6) = -1/6 x

... slope-intercept ... y = -1/6 x + 6

M ... point-slope ... (y + 1) = 2/3 (x + 3)
... slope-intercept ... y = 2/3 x + 1

substitute to find x ... 5 = 5/6 x

substitute back to find y