Question

name
4-3 skills practice
date
period
point - slope form
write an equation in point - slope form for the line that passes through the given point with the slope provided.
1.
2.
3.

1. (3, 1), m = 0
2. (-4, 6), m = 8
3. (1, -3), m = -4
4. (4, -6), m = 1
5. (3, 3), m=\frac{4}{3}
6. (-5, -1), m =-\frac{5}{4}

write each equation in standard form.

1. y + 1 = x + 2
2. y + 9 = -3(x - 2)
3. y - 7 = 4(x + 4)
4. y - 4 = -(x - 1)
5. y - 6 = 4(x + 3)
6. y + 5 = -5(x - 3)
7. y - 10 = -2(x - 3)
8. y - 2 =-\frac{1}{2}(x - 4)
9. y + 11 =\frac{1}{3}(x + 3)

write each equation in slope - intercept form.

1. y - 4 = 3(x - 2)
2. y + 2 = -(x + 4)
3. y - 6 = -2(x + 2)
4. y + 1 = -5(x - 3)
5. y - 3 = 6(x - 1)
6. y - 8 = 3(x + 5)
7. y - 2 =\frac{1}{2}(x + 6)
8. y + 1 =-\frac{1}{3}(x + 9)
9. y-\frac{1}{2}=x+\frac{1}{2}

glencoe algebra 1

Answer:

Let's solve these problems one by one. We'll start with the point - slope form questions, then move on to standard form and slope - intercept form.

Part 1: Write an equation in point - slope form

The point - slope form of a line is given by the formula \(y - y_1=m(x - x_1)\), where \((x_1,y_1)\) is a point on the line and \(m\) is the slope of the line.

Problem 1:

We are given the point \((-1,-2)\) and \(m = 3\).
Using the point - slope formula \(y - y_1=m(x - x_1)\), substitute \(x_1=-1\), \(y_1 = - 2\) and \(m = 3\).
We get \(y-(-2)=3(x - (-1))\), which simplifies to \(y + 2=3(x + 1)\).

Problem 2:

The point is \((1,-2)\) and \(m=-1\).
Substitute into the point - slope formula: \(y-(-2)=-1(x - 1)\), so \(y + 2=-(x - 1)\).

Problem 3:

The point is \((2,-3)\) and \(m = 0\).
Substitute into the formula: \(y-(-3)=0(x - 2)\), which simplifies to \(y + 3=0(x - 2)\) or \(y+3 = 0\) (but in point - slope form it is \(y+3=0(x - 2)\)).

Problem 4:

The point is \((3,1)\) and \(m = 0\).
Using the formula \(y - 1=0(x - 3)\).

Problem 5:

The point is \((-4,6)\) and \(m = 8\).
Substitute into the formula: \(y - 6=8(x+4)\) (since \(x_1=-4\), \(x - x_1=x-(-4)=x + 4\)).

Problem 6:

The point is \((1,-3)\) and \(m=-4\).
Substitute into the formula: \(y-(-3)=-4(x - 1)\), so \(y + 3=-4(x - 1)\).

Problem 7:

The point is \((4,-6)\) and \(m = 1\).
Substitute into the formula: \(y-(-6)=1(x - 4)\), so \(y + 6=x - 4\) (or in point - slope form \(y+6=(x - 4)\)).

Problem 8:

The point is \((3,3)\) and \(m=\frac{4}{3}\).
Substitute into the formula: \(y - 3=\frac{4}{3}(x - 3)\).

Problem 9:

The point is \((-5,-1)\) and \(m=-\frac{5}{4}\).
Substitute into the formula: \(y-(-1)=-\frac{5}{4}(x+5)\), so \(y + 1=-\frac{5}{4}(x + 5)\).

Part 2: Write each equation in standard form

The standard form of a linear equation is \(Ax+By = C\), where \(A\), \(B\), and \(C\) are integers and \(A\geq0\).

Problem 10:

We have the equation \(y + 1=x + 2\).
Subtract \(x\) from both sides: \(-x+y+1 = 2\).
Subtract 1 from both sides: \(-x+y=1\).
Multiply both sides by \(- 1\) to make \(A\geq0\): \(x - y=-1\).

Problem 11:

We have the equation \(y + 9=-3(x - 2)\).
First, expand the right - hand side: \(y + 9=-3x+6\).
Add \(3x\) to both sides: \(3x+y+9 = 6\).
Subtract 9 from both sides: \(3x+y=-3\).

Problem 12:

We have the equation \(y - 7=4(x + 4)\).
Expand the right - hand side: \(y - 7=4x+16\).
Subtract \(4x\) from both sides: \(-4x+y-7 = 16\).
Add 7 to both sides: \(-4x+y=23\).
Multiply both sides by \(-1\): \(4x - y=-23\).

Problem 13:

We have the equation \(y - 4=-(x - 1)\).
Expand the right - hand side: \(y - 4=-x + 1\).
Add \(x\) to both sides: \(x+y-4 = 1\).
Add 4 to both sides: \(x+y=5\).

Problem 14:

We have the equation \(y - 6=4(x + 3)\).
Expand the right - hand side: \(y - 6=4x+12\).
Subtract \(4x\) from both sides: \(-4x+y-6 = 12\).
Add 6 to both sides: \(-4x+y=18\).
Multiply both sides by \(-1\): \(4x - y=-18\).

Problem 15:

We have the equation \(y + 5=-5(x - 3)\).
Expand the right - hand side: \(y + 5=-5x+15\).
Add \(5x\) to both sides: \(5x+y+5 = 15\).
Subtract 5 from both sides: \(5x+y=10\).

Problem 16:

We have the equation \(y - 10=-2(x - 3)\).
Expand the right - hand side: \(y - 10=-2x+6\).
Add \(2x\) to both sides: \(2x+y-10 = 6\).
Add 10 to both sides: \(2x+y=16\).

Problem 17:

We have the equation \(y - 2=-\frac{1}{2}(x - 4)\).
Multiply both sides by 2 to eliminate the fraction: \(2(y - 2)=-(x - 4)\).
Expand both sides: \(2y-4=-x + 4\).
Add \(x\) to both sides: \(x+2y-4 = 4\).
Add 4 to both sides: \(x+2y=8\).

Problem 18:

We have the equation \(y + 11=\frac{1}{3}(x + 3)\).
Multiply both sides by 3 to eliminate the fraction: \(3(y + 11)=x + 3\).
Expand the left - hand side: \(3y+33=x + 3\).
Subtract \(x\) from both sides: \(-x+3y+33 = 3\).
Subtract 33 from both sides: \(-x+3y=-30\).
Multiply both sides by \(-1\): \(x - 3y=30\).

Part 3: Write each equation in slope - intercept form

The slope - intercept form of a line is \(y=mx + b\), where \(m\) is the slope and \(b\) is the y - intercept.

Problem 19:

We have the equation \(y - 4=3(x - 2)\).
Expand the right - hand side: \(y - 4=3x-6\).
Add 4 to both sides: \(y=3x-6 + 4\), so \(y=3x-2\).

Problem 20:

We have the equation \(y + 2=-(x + 4)\).
Expand the right - hand side: \(y + 2=-x-4\).
Subtract 2 from both sides: \(y=-x-4 - 2\), so \(y=-x-6\).

Problem 21:

We have the equation \(y - 6=-2(x + 2)\).
Expand the right - hand side: \(y - 6=-2x-4\).
Add 6 to both sides: \(y=-2x-4 + 6\), so \(y=-2x+2\).

Problem 22:

We have the equation \(y + 1=-5(x - 3)\).
Expand the right - hand side: \(y + 1=-5x+15\).
Subtract 1 from both sides: \(y=-5x+15 - 1\), so \(y=-5x+14\).

Problem 23:

We have the equation \(y - 3=6(x - 1)\).
Expand the right - hand side: \(y - 3=6x-6\).
Add 3 to both sides: \(y=6x-6 + 3\), so \(y=6x-3\).

Problem 24:

We have the equation \(y - 8=3(x + 5)\).
Expand the right - hand side: \(y - 8=3x+15\).
Add 8 to both sides: \(y=3x+15 + 8\), so \(y=3x+23\).

Problem 25:

We have the equation \(y - 2=\frac{1}{2}(x + 6)\).
Expand the right - hand side: \(y - 2=\frac{1}{2}x+3\).
Add 2 to both sides: \(y=\frac{1}{2}x+3 + 2\), so \(y=\frac{1}{2}x+5\).

Problem 26:

We have the equation \(y + 1=-\frac{1}{3}(x + 9)\).
Expand the right - hand side: \(y + 1=-\frac{1}{3}x-3\).
Subtract 1 from both sides: \(y=-\frac{1}{3}x-3 - 1\), so \(y=-\frac{1}{3}x-4\).

Problem 27:

We have the equation \(y-\frac{1}{2}=x+\frac{1}{2}\).
Add \(\frac{1}{2}\) to both sides: \(y=x+\frac{1}{2}+\frac{1}{2}\), so \(y=x + 1\).

Final Answers

Point - Slope Form:

1. \(y + 2=3(x + 1)\)
2. \(y + 2=-(x - 1)\)
3. \(y + 3=0(x - 2)\)
4. \(y - 1=0(x - 3)\)
5. \(y - 6=8(x + 4)\)
6. \(y + 3=-4(x - 1)\)
7. \(y + 6=x - 4\)
8. \(y - 3=\frac{4}{3}(x - 3)\)
9. \(y + 1=-\frac{5}{4}(x + 5)\)

Standard Form:

1. \(x - y=-1\)
2. \(3x + y=-3\)
3. \(4x - y=-23\)
4. \(x + y=5\)
5. \(4x - y=-18\)
6. \(5x + y=10\)
7. \(2x + y=16\)
8. \(x + 2y=8\)
9. \(x - 3y=30\)

Slope - Intercept Form:

1. \(y=3x - 2\)
2. \(y=-x - 6\)
3. \(y=-2x + 2\)
4. \(y=-5x + 14\)
5. \(y=6x - 3\)
6. \(y=3x + 23\)
7. \(y=\frac{1}{2}x+5\)
8. \(y=-\frac{1}{3}x - 4\)
9. \(y=x + 1\)