the straight L has gradient 3 and passes through the point A(-6,4).

a) find an equation for L in the form y=mx+c
b) straight line R has the equation in the form of -7y+x+14=0. Given that R crosses the y-axis at the point B and intersects L at the point C, find the coordinates of B and C

a) Let's use the point-slope form of a line to find the equation of L:

y - y₁ = m(x - x₁), where (x₁, y₁) is a point on the line and m is the gradient.

Substituting the given values, we have:
y - 4 = 3(x + 6)

Simplifying:
y - 4 = 3x + 18
y = 3x + 22

So the equation of line L is y = 3x + 22.

b) To find the coordinates of point B where R crosses the y-axis, we substitute x = 0 into the equation of line R:
-7y + 0 + 14 = 0
-7y = -14
y = 2

Therefore, point B is (0, 2).

To find the coordinates of point C where L intersects R, we set the equations of L and R equal to each other and solve for x and y:

3x + 22 = -7y + x + 14

Rearranging:
-7y - 3x = 8

Substituting y = 2, we can solve for x:
-7(2) - 3x = 8
-14 - 3x = 8
-3x = 8 + 14
-3x = 22
x = -22/3 ≈ -7.33

Now we substitute the value of x back into the equation of L to find y:
y = 3(-7.33) + 22
y ≈ -0.99

Therefore, point C is approximately (-7.33, -0.99).

So, the coordinates of point B are (0, 2), and the coordinates of point C are approximately (-7.33, -0.99).

To find the equation of the straight line L, we can use the point-slope form of a linear equation. The equation of a straight line can be written as y = mx + c, where m is the gradient (slope) of the line, and c is the y-intercept.

a) The gradient of L is given as 3, and it passes through the point A(-6, 4). Let's substitute the values into the equation y = mx + c:

4 = 3(-6) + c

Simplifying the equation:

4 = -18 + c

Adding 18 to both sides:

22 = c

Therefore, the equation of line L is y = 3x + 22.

b) The equation of line R is given as -7y + x + 14 = 0. We can rewrite it in the form y = mx + c:

-7y = -x - 14
y = (1/7)x + 2

To find the coordinates of point B, we need to find the y-intercept. When x = 0, y = 2. So, point B is (0, 2).

Now, we need to find the coordinates of point C, where lines L and R intersect. To do this, we can set the equations of L and R equal to each other and solve for x:

3x + 22 = (1/7)x + 2

Multiplying both sides by 7 to eliminate the fraction:

21x + 154 = x + 14

Subtracting x from both sides:

20x + 154 = 14

Subtracting 154 from both sides:

20x = -140

Dividing both sides by 20:

x = -7

Plugging x = -7 into the equation of line R to find y:

y = (1/7)(-7) + 2
y = -1 + 2
y = 1

Therefore, the coordinates of point C are (-7, 1).

a) To find the equation of the straight line L in the form y = mx + c, we need the gradient (m) and a point on the line (A(-6,4)).

1. Start with the equation y = mx + c.
2. Substitute the given gradient (m = 3) into the equation. Now we have y = 3x + c.
3. Use the coordinates of point A (-6,4) to find the value of c.
- Substitute x = -6 and y = 4 into the equation: 4 = 3(-6) + c.
- Simplify: 4 = -18 + c.
- Add 18 to both sides: 4 + 18 = c.
- Simplify: c = 22.
4. Substitute the value of c into the equation. The final equation for line L is y = 3x + 22.

b) To find the coordinates of point B and point C, we need to solve the equations of lines L and R.

1. Start with the given equation of line R: -7y + x + 14 = 0.
2. Rewrite the equation in the form y = mx + c:
-7y = -x - 14.
Divide by -7 to isolate y: y = (1/7)x + 2.
Now we have the equation of line R as y = (1/7)x + 2.
3. Find the y-intercept point B, where x = 0, by substituting x = 0 into the equation:
y = (1/7)(0) + 2.
Simplify: y = 2.
So point B is (0, 2).
4. Now we need to find the point of intersection between lines L and R, which is point C.
Set the equations of lines L and R equal to each other: 3x + 22 = (1/7)x + 2.
Subtract (1/7)x from both sides: 3x - (1/7)x + 22 = 2.
Combine like terms: (20/7)x + 22 = 2.
Subtract 22 from both sides: (20/7)x = -20.
Multiply both sides by 7/20 to isolate x: x = -7.
Substitute x = -7 into either equation to find y:
y = (1/7)(-7) + 2.
Simplify: y = -1 + 2.
So point C is (-7, 1).

Therefore, the coordinates of point B are (0, 2) and the coordinates of point C are (-7, 1).

equation for L:

y = 3x + b, but (-6,4) lies on it, so
4 = 3(-6) + b
b = 22
equation: y = 3x + 22

y-intercept of x-7y+14=0
let x = 0
-7y = -14
y = 2
so B is (0,2)

intersection of L and R:
sub L into R
x - 7y = -14
x - (3x+22) = -14
-2x = 8
x = -4 , then y = -12+22=10

so point C is (10,-4)

check my arithmetic.