1. Find the equation of the line described in each case:

a) Gradient 3, cuts through (0, 6).

b) Steepness of 4, passes through (0, -2).

c)Parallel to y = 5x - 1, cuts the y-axis at (0, 10).

d)Cuts through (0, 7) and (5, 17)

2. Here are 3 eqautions:
A. y + 1 = 3x
B. y = -3(x - 1)
C. 2y = 8x + 6

Find the odd one out in each case:
a) Gradient is ±3
b) Slopes in a positive direction
c) Cuts the y-axis at (0, 3)

3. A line passes through the point (10, 10). It's equation is y = mx + 4. Find the value of m.

I'll try to get you started on these, and let you take it from there.

For 1a:
y - y1 = m(x - x1)
y - 6 = 3(x - 0)
y - 6 = 3x
y = 3x + 6

For 1b:
Use the same procedure as 1a.

For 1c:
Parallel lines have the same slope.
y - 10 = 5(x - 0)
I'll let you finish.

For 1d:
m = (y2 - y1)/(x2 - x1)
m = (17 - 7)/(5 - 0)
m = 10/5 = 2
Therefore, using the first ordered pair (0, 7), we have this:
(y - 7)/(x - 0) = 2
y - 7 = 2(x - 0)
I'll let you take it from here to finish.

For 2a:
The coefficient of x is the slope.

For 2b:
Look for positive slopes.

For 2c:
Substitute 0 for x and 3 for y in each equation to check each one.

For 3:
Substitute 10 for x and 10 for y. Solve the equation for m, which is the slope.

I hope this will help get you started on these problems.

<<3. A line passes through the point (10, 10). It's equation is y = mx + 4. Find the value of m>>
Since y = mx + 4 and y = 10 when x = 10, plug those values of x and y into the eqwuation and solve for m.

Make an attempt at the others and we will glad to critique your work. Please limit your questions to one per post.

That's a lot of equations! Let's start with the first one.

1a) To find the equation of a line with a gradient of 3 and passing through (0, 6), we can use the point-slope form of the equation: y - y1 = m(x - x1), where m is the gradient and (x1, y1) is a point on the line.

So, substituting the values into the equation, we get:
y - 6 = 3(x - 0)
y - 6 = 3x
y = 3x + 6

1b) Similarly, for a line with a steepness of 4 and passing through (0, -2), we have:
y - (-2) = 4(x - 0)
y + 2 = 4x
y = 4x - 2

1c) To find the equation of a line parallel to y = 5x - 1 and passing through (0, 10), we know that parallel lines have the same slope. So the equation for the line will also have a slope of 5. Using the point-slope equation again, we have:
y - 10 = 5(x - 0)
y - 10 = 5x
y = 5x + 10

1d) To find the equation of a line passing through (0, 7) and (5, 17), we can use the slope-intercept form of the equation: y = mx + b, where m is the slope and b is the y-intercept.

First, let's find the slope:
m = (y2 - y1) / (x2 - x1)
m = (17 - 7) / (5 - 0)
m = 10 / 5
m = 2

Now, substituting the values into the equation, we have:
y = 2x + b

Using the point (0, 7), we can solve for b:
7 = 2(0) + b
7 = b

So, the equation of the line is:
y = 2x + 7

Now, let's move on to the second set of questions.

2a) To find the odd one out in terms of the gradient of ±3, let's check each equation.

- Equation A: y + 1 = 3x
The gradient is 3.

- Equation B: y = -3(x - 1)
The gradient is -3.

- Equation C: 2y = 8x + 6
Dividing by 2 gives us y = 4x + 3.
The gradient is 4.

So, the odd one out in terms of a gradient of ±3 is Equation C.

2b) To find the odd one out in terms of slopes in a positive direction, let's check each equation.

- Equation A: y + 1 = 3x
The slope is positive (3).

- Equation B: y = -3(x - 1)
The slope is negative (-3).

- Equation C: 2y = 8x + 6
Dividing by 2 gives us y = 4x + 3.
The slope is positive (4).

So, the odd one out in terms of slopes in a positive direction is Equation B.

2c) To find the equation that cuts the y-axis at (0, 3), let's check each equation by substituting x = 0 and y = 3.

- Equation A: y + 1 = 3x
Substituting x = 0 and y = 3 gives us 3 + 1 = 3(0), which is not true.

- Equation B: y = -3(x - 1)
Substituting x = 0 and y = 3 gives us 3 = -3(0 - 1), which is true.

- Equation C: 2y = 8x + 6
Substituting x = 0 and y = 3 gives us 2(3) = 8(0) + 6, which is not true.

So, the odd one out in terms of cutting the y-axis at (0, 3) is Equation A.

Lastly, for question 3, we are given that the line passes through the point (10, 10) and its equation is y = mx + 4. We need to find the value of m.

Substituting the given point into the equation, we have:
10 = m(10) + 4
10 = 10m + 4
6 = 10m
m = 6/10
m = 3/5

Therefore, the value of m is 3/5.

1a) To find the equation of a line with a gradient of 3 and passing through the point (0,6), we can use the point-slope form of the equation.

The point-slope form is given by y - y1 = m(x - x1), where (x1, y1) is the given point and m is the gradient.
Substituting the values, we have y - 6 = 3(x - 0).
Simplifying, we get y - 6 = 3x.
Therefore, the equation of the line is y = 3x + 6.

1b) Similarly, for a line with a steepness of 4 and passing through the point (0,-2), we can again use the point-slope form.
We have y - (-2) = 4(x - 0).
Simplifying, we get y + 2 = 4x.
Therefore, the equation of the line is y = 4x - 2.

1c) To find the equation of a line parallel to y = 5x - 1 and passing through the point (0,10), we can use the fact that parallel lines have the same gradient.
Since y = 5x - 1 has a gradient of 5, the parallel line will also have a gradient of 5.
Using the point-slope form, we have y - 10 = 5(x - 0).
Simplifying, we get y - 10 = 5x.
Therefore, the equation of the line is y = 5x + 10.

1d) To find the equation of a line passing through the points (0,7) and (5,17), we can use the slope-intercept form of the equation.
The slope-intercept form is given by y = mx + c, where m is the gradient and c is the y-intercept.
First, find the gradient using the formula: m = (y2 - y1)/(x2 - x1).
Substituting the values, we have m = (17 - 7)/(5 - 0) = 10/5 = 2.
Next, choose one of the points (0,7) and plug in the values of x, y, and m.
We have 7 = 2(0) + c.
Simplifying, we get c = 7.
Therefore, the equation of the line is y = 2x + 7.

2a) To determine which equation has a gradient of ±3, we can compare the coefficient of x in each equation.
In equation A: y + 1 = 3x, the coefficient of x is 3, which matches the desired gradient.
In equation B: y = -3(x - 1), the coefficient of x is -3, which matches the desired gradient.
In equation C: 2y = 8x + 6, the coefficient of x is 8, which does not match the desired gradient of ±3. Therefore, the odd one out is equation C.

2b) To determine which equation has slopes in a positive direction, we can compare the sign of the coefficient of x in each equation.
In equation A: y + 1 = 3x, the coefficient of x is 3, which is positive.
In equation B: y = -3(x - 1), the coefficient of x is -3, which is negative.
In equation C: 2y = 8x + 6, the coefficient of x is 8, which is positive. Therefore, the odd one out is equation B.

2c) To determine which equation cuts the y-axis at (0,3), we can substitute x = 0 and y = 3 into each equation and check if it satisfies the condition.
In equation A: 3 + 1 = 3(0), which is not true.
In equation B: 3 = -3(0 - 1), which is true.
In equation C: 2(3) = 8(0) + 6, which is not true. Therefore, the odd one out is equation A.

3. Given that the equation of the line passing through the point (10,10) is y = mx + 4, we can substitute the values of x and y into the equation and solve for m.
We have 10 = m(10) + 4.
Subtracting 4 from both sides, we get 6 = m(10).
Dividing both sides by 10, we get m = 6/10 = 3/5.
Therefore, the value of m is 3/5.

Sure! Let's go through each question one by one.

1a) To find the equation of a line with gradient 3 and passing through (0, 6), we can use the point-slope form of a line.
The equation is given by: y - y1 = m(x - x1), where (x1, y1) represents the given point and m represents the gradient.
Plugging in the values, we have: y - 6 = 3(x - 0)
Simplifying the equation, we get: y - 6 = 3x
Rearranging, we find the equation of the line: y = 3x + 6

1b) Similar to the previous question, for a line with a steepness of 4 and passing through (0, -2), we use the point-slope form:
y - y1 = m(x - x1).
Substituting the values, we get: y - (-2) = 4(x - 0)
Simplifying the equation, we have: y + 2 = 4x
Rearranging, we find the equation: y = 4x - 2

1c) To find the equation of a line parallel to y = 5x - 1 and passing through (0, 10), we know that parallel lines have the same gradient.
So the new line also has a gradient of 5. Plugging in the values, we have: y - 10 = 5(x - 0)
Simplifying the equation, we get: y - 10 = 5x
Rearranging, we find the equation: y = 5x + 10

1d) To find the equation of a line passing through (0, 7) and (5, 17), we can use the slope-intercept form of a line, which is y = mx + b.
First, calculate the gradient. We have: m = (y2 - y1)/(x2 - x1) = (17 - 7)/(5 - 0) = 10/5 = 2.
Using the first point (0, 7), we can substitute the values into the slope-intercept form:
y - 7 = 2(x - 0)
Simplifying the equation, we get: y - 7 = 2x
Rearranging, we find the equation: y = 2x + 7

2a) To find the odd one out based on gradient ±3, we check the coefficient of x in each equation.
Equation A: y + 1 = 3x, the coefficient of x is 3.
Equation B: y = -3(x - 1), the coefficient of x is -3.
Equation C: 2y = 8x + 6, the coefficient of x is 8/2 = 4.
The odd one out is Equation C because it does not have a gradient of ±3.

2b) To find the odd one out based on slopes in a positive direction, we check the sign of the coefficient of x in each equation.
Equation A: y + 1 = 3x, the slope is positive (3x).
Equation B: y = -3(x - 1), the slope is negative (-3x).
Equation C: 2y = 8x + 6, the slope is positive (8x).
The odd one out is Equation B because it has a negative slope.

2c) To find the odd one out based on cutting the y-axis at (0, 3), we substitute x = 0 and y = 3 into each equation.
Equation A: 3 + 1 = 3(0), 4 = 0, which is false.
Equation B: 3 = -3(0 - 1), 3 = -3, which is false.
Equation C: 2(3) = 8(0) + 6, 6 = 6, which is true.
The odd one out is Equation A because it does not satisfy the condition of cutting the y-axis at (0, 3).

3. To find the value of m in the equation y = mx + 4, we substitute x = 10 and y = 10.
Plugging the values, we have: 10 = m(10) + 4
Solving the equation, we get: 10 - 4 = 10m
6 = 10m
Dividing both sides by 10, we find: m = 6/10 = 0.6

I hope this helps! Let me know if you have any further questions.