SATPREP assignment focusing on coordinate geometry, covering gradient calculations, line equations, and point verification.

Let’s solve each problem step by step. I’ll go through them one at a time, showing all the work so you can follow along easily.

---

Problem 1: Find the gradient of the line segment joining each pair of points.

Gradient formula:
> m = (y₂ - y₁) / (x₂ - x₁)

We’ll apply this to each pair.

a) (3, 1) and (5, 5)
m = (5 - 1)/(5 - 3) = 4/2 = 2

b) (4, 7) and (10, 9)
m = (9 - 7)/(10 - 4) = 2/6 = 1/3

c) (6, 1) and (2, 5)
m = (5 - 1)/(2 - 6) = 4/(-4) = -1

d) (-2, 2) and (2, 8)
m = (8 - 2)/(2 - (-2)) = 6/4 = 3/2

e) (1, 3) and (7, -1)
m = (-1 - 3)/(7 - 1) = (-4)/6 = -2/3

f) (4, 5) and (-5, -7)
m = (-7 - 5)/(-5 - 4) = (-12)/(-9) = 4/3

g) (-2, 0) and (0, -8)
m = (-8 - 0)/(0 - (-2)) = (-8)/2 = -4

h) (8, 6) and (-7, -2)
m = (-2 - 6)/(-7 - 8) = (-8)/(-15) = 8/15

✔ All gradients calculated.

---

Problem 2: Write down the gradient and y-intercept of each line.

These are already in slope-intercept form: y = mx + c → gradient is m, y-intercept is c.

a) y = 4x - 1 → gradient = 4, y-intercept = -1

b) y = (1/3)x + 3 → gradient = 1/3, y-intercept = 3

c) y = 6 - x → rewrite as y = -x + 6 → gradient = -1, y-intercept = 6

d) y = -2x - 3/5 → gradient = -2, y-intercept = -3/5

✔ Done.

---

Problem 3: Find the gradient and y-intercept of each line.

Need to rearrange into y = mx + c form.

a) x + y + 3 = 0 → y = -x - 3 → gradient = -1, y-intercept = -3

b) x - 2y - 6 = 0 → -2y = -x + 6 → y = (1/2)x - 3 → gradient = 1/2, y-intercept = -3

c) 3x + 3y - 2 = 0 → 3y = -3x + 2 → y = -x + 2/3 → gradient = -1, y-intercept = 2/3

d) 4x - 5y + 1 = 0 → -5y = -4x - 1 → y = (4/5)x + 1/5 → gradient = 4/5, y-intercept = 1/5

✔ Done.

---

Problem 4: Write equation in point-slope form: y - y₁ = m(x - x₁)

Just plug in given m and point (x₁, y₁).

a) m=2, point (4,1) → y - 1 = 2(x - 4)

b) m=5, point (2,-5) → y - (-5) = 5(x - 2) → y + 5 = 5(x - 2)

c) m=-3, point (-1,1) → y - 1 = -3(x - (-1)) → y - 1 = -3(x + 1)

d) m=1/2, point (1,6) → y - 6 = (1/2)(x - 1)

e) m=-2, point (3/4, -1/4) → y - (-1/4) = -2(x - 3/4) → y + 1/4 = -2(x - 3/4)

f) m=-1/5, point (-3,-7) → y - (-7) = (-1/5)(x - (-3)) → y + 7 = (-1/5)(x + 3)

✔ All done.

---

Problem 5: Find equation in y = mx + c form, given gradient and point.

Use y = mx + c, plug in m and point to find c.

a) m=3, point (1,2):
2 = 3(1) + c → 2 = 3 + c → c = -1 → y = 3x - 1

b) m=-1, point (5,3):
3 = -1(5) + c → 3 = -5 + c → c = 8 → y = -x + 8

c) m=4, point (-2,-3):
-3 = 4(-2) + c → -3 = -8 + c → c = 5 → y = 4x + 5

d) m=-2, point (-4,1):
1 = -2(-4) + c → 1 = 8 + c → c = -7 → y = -2x - 7

e) m=1/3, point (-3,1):
1 = (1/3)(-3) + c → 1 = -1 + c → c = 2 → y = (1/3)x + 2

f) m=-5/6, point (9,-2):
-2 = (-5/6)(9) + c → -2 = -45/6 + c → -2 = -7.5 + c → c = 5.5 = 11/2
→ y = (-5/6)x + 11/2

✔ All equations found.

---

Problem 6: Equation in ax + by + c = 0 form, with integer coefficients.

Start from point-slope or slope-intercept, then rearrange.

a) m=1, P(2,-4)
Point-slope: y - (-4) = 1(x - 2) → y + 4 = x - 2 → x - y - 6 = 0 → x - y - 6 = 0

b) m=1/2, P(6,1)
y - 1 = (1/2)(x - 6) → multiply both sides by 2: 2y - 2 = x - 6 → x - 2y - 4 = 0 → x - 2y - 4 = 0

c) m=-4, P(-1,8)
y - 8 = -4(x + 1) → y - 8 = -4x - 4 → 4x + y - 4 = 0 → 4x + y - 4 = 0

d) m=2/3, P(-3,5)
y - 5 = (2/3)(x + 3) → multiply by 3: 3y - 15 = 2x + 6 → 2x - 3y + 21 = 0? Wait — let's check:

3y - 15 = 2x + 6 → bring all to left: -2x + 3y - 21 = 0 → multiply by -1: 2x - 3y + 21 = 0

Wait — actually, standard is usually positive leading coefficient. Let me recompute:

From y - 5 = (2/3)(x + 3)

Multiply both sides by 3: 3(y - 5) = 2(x + 3) → 3y - 15 = 2x + 6

Bring all to left: -2x + 3y - 21 = 0 → multiply by -1: 2x - 3y + 21 = 0

But let’s verify with point: plug x=-3, y=5 → 2(-3) -3(5) +21 = -6 -15 +21 = 0 ✔

e) m=-3, P(3/2, -1/8)
y - (-1/8) = -3(x - 3/2) → y + 1/8 = -3x + 9/2

Multiply entire equation by 8 to eliminate denominators:

8y + 1 = -24x + 36 → 24x + 8y - 35 = 0 → 24x + 8y - 35 = 0

Check: x=3/2, y=-1/8 → 24*(3/2)=36, 8*(-1/8)=-1 → 36 -1 -35=0 ✔

f) m=-3/4, P(2/3, -7)
y - (-7) = (-3/4)(x - 2/3) → y + 7 = (-3/4)x + (3/4)*(2/3) = (-3/4)x + 1/2

Multiply entire equation by 4: 4y + 28 = -3x + 2 → 3x + 4y + 26 = 0 → 3x + 4y + 26 = 0

Check: x=2/3, y=-7 → 3*(2/3)=2, 4*(-7)=-28 → 2 -28 +26=0 ✔

✔ All done.

---

Problem 7: Find equation in y = mx + c form through two points.

First find gradient using m = (y₂-y₁)/(x₂-x₁), then use one point to find c.

a) (0,1) and (4,13)
m = (13-1)/(4-0) = 12/4 = 3
Using (0,1): y = 3x + c → 1 = 0 + c → c=1 → y = 3x + 1

b) (2,9) and (7,-1)
m = (-1-9)/(7-2) = -10/5 = -2
Using (2,9): 9 = -2(2) + c → 9 = -4 + c → c=13 → y = -2x + 13

c) (-4,3) and (2,7)
m = (7-3)/(2+4) = 4/6 = 2/3
Using (2,7): 7 = (2/3)(2) + c → 7 = 4/3 + c → c = 7 - 4/3 = 17/3 → y = (2/3)x + 17/3

d) (-1/2, -2) and (2,8)
m = (8 - (-2))/(2 - (-1/2)) = 10 / (2.5) = 10 / (5/2) = 10 * 2/5 = 4
Using (2,8): 8 = 4(2) + c → 8 = 8 + c → c=0 → y = 4x

e) (3,-2) and (18,-5)
m = (-5 - (-2))/(18 - 3) = (-3)/15 = -1/5
Using (3,-2): -2 = (-1/5)(3) + c → -2 = -3/5 + c → c = -2 + 3/5 = -7/5 → y = (-1/5)x - 7/5

f) (-3.2, 4) and (-2, 0.4)
Convert to fractions for accuracy:
-3.2 = -16/5, 0.4 = 2/5
Points: (-16/5, 4) and (-2, 2/5) = (-10/5, 2/5)

m = (2/5 - 4) / (-10/5 + 16/5) = (2/5 - 20/5) / (6/5) = (-18/5) / (6/5) = -18/6 = -3

Now use point (-2, 0.4): y = -3x + c → 0.4 = -3(-2) + c → 0.4 = 6 + c → c = -5.6 = -28/5

So y = -3x - 28/5

Or decimal: y = -3x - 5.6

But since original had decimals, maybe keep decimal? But question doesn’t specify. Let’s write as fraction.

Actually, better to use exact values.

Original points: (-3.2, 4) and (-2, 0.4)

m = (0.4 - 4)/(-2 - (-3.2)) = (-3.6)/(1.2) = -3 → same.

Then y = -3x + c → plug in (-2, 0.4): 0.4 = -3*(-2) + c → 0.4 = 6 + c → c = -5.6

So y = -3x - 5.6

But if we want fraction: -5.6 = -28/5 → y = -3x - 28/5

Either is fine, but since input had decimals, perhaps decimal is acceptable. However, in math problems, fractions are preferred unless specified.

I’ll go with fraction: y = -3x - 28/5

✔ Done.

---

Problem 8: Equation in ax + by + c = 0 form, integers, through two points.

Find gradient first, then use point-slope, then rearrange to standard form with integers.

a) (3,0) and (5,2)
m = (2-0)/(5-3) = 2/2 = 1
Point-slope: y - 0 = 1(x - 3) → y = x - 3 → x - y - 3 = 0 → x - y - 3 = 0

b) (-1,8) and (5,-4)
m = (-4-8)/(5+1) = -12/6 = -2
Using (-1,8): y - 8 = -2(x + 1) → y - 8 = -2x - 2 → 2x + y - 6 = 0 → 2x + y - 6 = 0

c) (-5,3) and (7,5)
m = (5-3)/(7+5) = 2/12 = 1/6
Using (-5,3): y - 3 = (1/6)(x + 5)
Multiply by 6: 6y - 18 = x + 5 → x - 6y + 23 = 0? Wait:

6y - 18 = x + 5 → bring all to left: -x + 6y - 23 = 0 → multiply by -1: x - 6y + 23 = 0

Check: x=-5,y=3 → -5 -18 +23=0 ✔; x=7,y=5 → 7 -30 +23=0 ✔

d) (-4,-1) and (8,-17)
m = (-17+1)/(8+4) = -16/12 = -4/3
Using (-4,-1): y +1 = (-4/3)(x +4)
Multiply by 3: 3y +3 = -4x -16 → 4x + 3y +19 = 0 → 4x + 3y + 19 = 0

Check: x=-4,y=-1 → 4*(-4)+3*(-1)+19 = -16-3+19=0 ✔; x=8,y=-17 → 32 -51 +19=0 ✔

e) (2,-1.5) and (7,0)
Write -1.5 as -3/2
Points: (2, -3/2), (7,0)
m = (0 - (-3/2))/(7-2) = (3/2)/5 = 3/10
Using (7,0): y - 0 = (3/10)(x - 7)
Multiply by 10: 10y = 3x - 21 → 3x - 10y - 21 = 0 → 3x - 10y - 21 = 0

Check: x=2,y=-1.5 → 3*2=6, -10*(-1.5)=15 → 6+15-21=0 ✔; x=7,y=0 → 21 -0 -21=0 ✔

f) (-3/5, 1/10) and (3,1)
m = (1 - 1/10)/(3 - (-3/5)) = (9/10)/(3 + 3/5) = (9/10)/(18/5) = (9/10)*(5/18) = 45/180 = 1/4

Using (3,1): y - 1 = (1/4)(x - 3)
Multiply by 4: 4y - 4 = x - 3 → x - 4y +1 = 0? Wait:

4y - 4 = x - 3 → bring all to left: -x + 4y -1 = 0 → multiply by -1: x - 4y + 1 = 0

Check: x=3,y=1 → 3 -4 +1=0 ✔; x=-3/5,y=1/10 → (-3/5) -4*(1/10) +1 = -3/5 - 2/5 +1 = -1 +1=0 ✔

✔ All done.

---

Problem 9: Line l passes through A(-6,8) and B(3,2)

a) Find equation of line l.

First, gradient m = (2 - 8)/(3 - (-6)) = (-6)/9 = -2/3

Use point A(-6,8): y - 8 = (-2/3)(x + 6)

Simplify: y - 8 = (-2/3)x - 4 → y = (-2/3)x + 4

Or in standard form: multiply by 3: 3y = -2x + 12 → 2x + 3y - 12 = 0

But question says “find an equation”, so either form is fine. Since part b asks to show point lies on it, let’s use y = (-2/3)x + 4

b) Show that C(9,-2) lies on l.

Plug x=9 into equation: y = (-2/3)(9) + 4 = -6 + 4 = -2 → which matches y-coordinate of C.

So yes, C lies on l.

✔ Done.

---

Problem 10: Point M(k, 2k) lies on line x - 3y + 15 = 0. Find k.

Substitute x=k, y=2k into equation:

k - 3(2k) + 15 = 0 → k - 6k + 15 = 0 → -5k + 15 = 0 → -5k = -15 → k = 3

✔ Check: x=3, y=6 → 3 - 18 + 15 = 0 ✔

---

Final Answer:

Here are all the answers compiled clearly:

1. Gradients:
a) 2
b) 1/3
c) -1
d) 3/2
e) -2/3
f) 4/3
g) -4
h) 8/15

2. Gradient and y-intercept:
a) grad=4, int=-1
b) grad=1/3, int=3
c) grad=-1, int=6
d) grad=-2, int=-3/5

3. Gradient and y-intercept:
a) grad=-1, int=-3
b) grad=1/2, int=-3
c) grad=-1, int=2/3
d) grad=4/5, int=1/5

4. Point-slope form:
a) y - 1 = 2(x - 4)
b) y + 5 = 5(x - 2)
c) y - 1 = -3(x + 1)
d) y - 6 = (1/2)(x - 1)
e) y + 1/4 = -2(x - 3/4)
f) y + 7 = (-1/5)(x + 3)

5. Slope-intercept form:
a) y = 3x - 1
b) y = -x + 8
c) y = 4x + 5
d) y = -2x - 7
e) y = (1/3)x + 2
f) y = (-5/6)x + 11/2

6. Standard form (ax+by+c=0, integers):
a) x - y - 6 = 0
b) x - 2y - 4 = 0
c) 4x + y - 4 = 0
d) 2x - 3y + 21 = 0
e) 24x + 8y - 35 = 0
f) 3x + 4y + 26 = 0

7. Slope-intercept form through two points:
a) y = 3x + 1
b) y = -2x + 13
c) y = (2/3)x + 17/3
d) y = 4x
e) y = (-1/5)x - 7/5
f) y = -3x - 28/5

8. Standard form through two points:
a) x - y - 3 = 0
b) 2x + y - 6 = 0
c) x - 6y + 23 = 0
d) 4x + 3y + 19 = 0
e) 3x - 10y - 21 = 0
f) x - 4y + 1 = 0

9.
a) y = (-2/3)x + 4 (or 2x + 3y - 12 = 0)
b) Substituting x=9 gives y=-2, so point C(9,-2) lies on the line.

10. k = 3