Linear Equation Practice Problems | Equation of Straight Line

Master finding linear equations using slope-intercept form, two points, parallel lines, and perpendicular lines with step-by-step practice problems and solutions.

πŸ“šMaster Linear Equation Methods with Targeted Practice
  • Find linear equations using slope and a given point with y=mx+b formula
  • Calculate slope from two points and determine the complete equation
  • Identify parallel line equations using equal slope relationships
  • Solve perpendicular line problems using m1Γ—m2=-1 rule
  • Read graphs to extract points and create linear equations
  • Apply point-slope form y-y1=m(x-x1) in various contexts

Understanding Equation of a Straight Line

Complete explanation with examples

Finding a linear equation is actually about graphing the linear function using y=mx+b y=mx+b or y=mx y=mx .

A - linear function using y=mx+b or y=mx

We can find the linear equation in 5 5 ways:

  • Using a point on the line and the slope of the line.
  • Using two points that lie on the line.
  • Using the graph of the function itself.
  • Using parallel lines, that is, if the requested line is parallel to another line and we know the slope of the other line.
  • Using perpendicular lines, that is, if the requested line is perpendicular to another line and we know the slope of the other line.

The first three methods are based in one way or another on the general formula for finding the linear equation:

yβˆ’y1=mΓ—(xβˆ’x1) y-y1=m\times\left(x-x1\right)

The last two methods also use this formula, but they also take into account two additional rules:

  • For parallel lines, the slopes are equal, that is m1=m2 m1=m2
  • For perpendicular lines, the slopes have the relationship m1Γ—m2=βˆ’1 m1\times m2=-1
Detailed explanation

Practice Equation of a Straight Line

Test your knowledge with 5 quizzes

Two lines haves slopes of -3 and -6.

Which of the lines forms a greater angle with the x axis?

Examples with solutions for Equation of a Straight Line

Step-by-step solutions included
Exercise #1

Find the equation of the line passing through the two points (βˆ’2,βˆ’6),(4,12) (-2,-6),(4,12)

Step-by-Step Solution

In the first step, we'll find the slope using the formula:

m=y2βˆ’y1x2βˆ’x1 m=\frac{y_2-y_1}{x_2-x_1}

We'll substitute according to the given points:

m=12βˆ’(βˆ’6)4βˆ’(βˆ’2) m=\frac{12-(-6)}{4-(-2)}

m=186=3 m=\frac{18}{6}=3

Now we'll choose the point (4,12) and use the formula:

y=mx+b y=mx+b

12=3Γ—4+b 12=3\times4+b

12=12+b 12=12+b

b=0 b=0

We'll substitute the data into the formula to find the equation of the line:

y=3x y=3x

Answer:

y=3x y=3x

Video Solution
Exercise #2

Find the equation of the line passing through the two points (13,1),(βˆ’13,2) (\frac{1}{3},1),(-\frac{1}{3},2)

Step-by-Step Solution

In the first step, we'll find the slope using the formula:

m=y2βˆ’y1x2βˆ’x1 m=\frac{y_2-y_1}{x_2-x_1}

We'll substitute according to the given points:

m=1βˆ’213βˆ’(βˆ’13) m=\frac{1-2}{\frac{1}{3}-(-\frac{1}{3})}

m=βˆ’123=βˆ’32 m=\frac{-1}{\frac{2}{3}}=-\frac{3}{2}

Now we'll choose point (13,1) (\frac{1}{3},1) and use the formula:

y=mx+b y=mx+b

1=βˆ’32Γ—13+b 1=-\frac{3}{2}\times\frac{1}{3}+b

1=βˆ’12+b 1=-\frac{1}{2}+b

b=112 b=1\frac{1}{2}

We'll substitute the given data into the formula to find the equation of the line:

y=βˆ’32x+112 y=-\frac{3}{2}x+1\frac{1}{2}

Answer:

y=βˆ’32x+112 y=-\frac{3}{2}x+1\frac{1}{2}

Video Solution
Exercise #3

Find the equation of the line passing through the two points (9,10),(99,100) (9,10),(99,100)

Step-by-Step Solution

To find the equation of the line passing through the points (9,10)(9, 10) and (99,100)(99, 100), follow these steps:

  • Step 1: Determine the slope, mm, of the line.
  • Step 2: Use one point and the calculated slope to find the equation of the line.
  • Step 3: Simplify to find the line's equation in slope-intercept form.

Step 1: Calculate the slope m m using the formula:
m=100βˆ’1099βˆ’9=9090=1 m = \frac{100 - 10}{99 - 9} = \frac{90}{90} = 1 .

Step 2: Now, use the point (9,10)(9, 10) and the point-slope form:
yβˆ’10=1β‹…(xβˆ’9) y - 10 = 1 \cdot (x - 9) .

Step 3: Simplify this equation:
yβˆ’10=xβˆ’9 y - 10 = x - 9
y=x+1 y = x + 1 .

Thus, the equation of the line is y=x+1 y = x + 1 , matching answer choice (2).

Answer:

y=x+1 y=x+1

Video Solution
Exercise #4

Find the equation of the line passing through the two points (2,8),(6,1) (2,8),(6,1)

Step-by-Step Solution

To find the equation of the line passing through the points (2,8) (2,8) and (6,1) (6,1) , follow the steps below:

  • Step 1: Calculate the slope m m .
    The formula for the slope m m is:

m=y2βˆ’y1x2βˆ’x1=1βˆ’86βˆ’2=βˆ’74 m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{1 - 8}{6 - 2} = \frac{-7}{4}

  • Step 2: Use the point-slope form to write the equation of the line.
    The point-slope form of a line is given by:

yβˆ’y1=m(xβˆ’x1) y - y_1 = m(x - x_1)

Using the point (2,8)(2,8):

yβˆ’8=βˆ’74(xβˆ’2) y - 8 = -\frac{7}{4}(x - 2)

  • Step 3: Simplify to the slope-intercept form.
    Distribute the slope and rearrange to find y y :

yβˆ’8=βˆ’74x+72 y - 8 = -\frac{7}{4}x + \frac{7}{2}

Add 8 to both sides to solve for y y :

y=βˆ’74x+72+8 y = -\frac{7}{4}x + \frac{7}{2} + 8

Convert 8 to a fraction with a denominator of 2:

y=βˆ’74x+72+162 y = -\frac{7}{4}x + \frac{7}{2} + \frac{16}{2}

Simplify the addition:

y=βˆ’74x+232 y = -\frac{7}{4}x + \frac{23}{2}

To convert 23/2 23/2 into mixed number form: 232=1112 \frac{23}{2} = 11 \frac{1}{2}

Thus, the equation in slope-intercept form is: y=βˆ’74x+1112 y = -\frac{7}{4}x + 11 \frac{1}{2}

Therefore, the equation of the line passing through these points is y=βˆ’134x+1112 y = -1\frac{3}{4}x + 11\frac{1}{2} , which matches the correct choice in the multiple-choice answers.

Answer:

y=βˆ’134x+1112 y=-1\frac{3}{4}x+11\frac{1}{2}

Video Solution
Exercise #5

Find the equation of the line passing through the two points (5,0),(12,412) (5,0),(\frac{1}{2},4\frac{1}{2})

Step-by-Step Solution

First, we will use the formula to find the slope of the straight line:

We replace the data and solve:

(0βˆ’4.5)(5βˆ’0.5)=βˆ’4.54.5=βˆ’1 \frac{(0-4.5)}{(5-0.5)}=\frac{-4.5}{4.5}=-1

Now, we know that the slope is βˆ’1 -1

Β 

We replace one of the points in the formula of the line equation:

y=mx+b y=mx+b

(5,0) (5,0)

0=βˆ’1Γ—5+b 0=-1\times5+b

Β 0=βˆ’5+b 0=-5+b

b=5 b=5

Now we have the data to complete the equation:

y=βˆ’1Γ—x+5 y=-1\times x+5

y=βˆ’x+5 y=-x+5

Answer:

y+x=5 y+x=5

Video Solution

Frequently Asked Questions

What are the 5 methods to find the equation of a straight line?

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The five methods are: 1) Using a point and slope, 2) Using two points on the line, 3) Using the graph of the function, 4) Using parallel lines (equal slopes), and 5) Using perpendicular lines (slopes multiply to -1). All methods ultimately use the point-slope form y-y1=m(x-x1).

How do you find a linear equation when given slope and one point?

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Substitute the given slope (m) and point coordinates into y=mx+b. Solve for b by plugging in the x and y values of the point. For example, with slope -2 and point (2,4): 4=-2(2)+b, so b=8, giving y=-2x+8.

What is the slope formula using two points?

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The slope formula is m=(Y2-Y1)/(X2-X1). Take the difference of y-coordinates divided by the difference of x-coordinates. Once you find the slope, use either point with the point-slope form to get the complete equation.

How do parallel lines help find linear equations?

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Parallel lines have identical slopes (m1=m2). If you know a line is parallel to y=3x-7, your line also has slope 3. Combined with a given point, you can find the y-intercept and complete equation using y=mx+b.

What is the relationship between perpendicular line slopes?

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Perpendicular lines have slopes that multiply to -1 (m1Γ—m2=-1). If one line has slope 2, the perpendicular line has slope -1/2. This relationship helps you find the slope when given a perpendicular reference line.

How do you find a linear equation from a graph?

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Choose any two clear points from the graph, calculate the slope using m=(Y2-Y1)/(X2-X1), then use one point with the slope in the point-slope form. This method combines the two-point method with visual information from the graph.

What is the difference between y=mx+b and point-slope form?

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y=mx+b is slope-intercept form showing slope (m) and y-intercept (b). Point-slope form is y-y1=m(x-x1) using a specific point (x1,y1) and slope. Point-slope form is more versatile for finding equations, while slope-intercept form is better for graphing.

Why do you need the equation to be explicit to find slope?

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The slope is the coefficient of x when y is isolated on one side with coefficient 1. In implicit forms like 2y=6x+4, you must first solve for y to get y=3x+2, revealing the slope is 3, not 6.

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