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:

yy1=m×(xx1) 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 m1×m2=1 m1\times m2=-1

Suggested Topics to Practice in Advance

  1. Function
  2. Linear Function
  3. The Linear Function y=mx+b
  4. Slope in the Function y=mx
  5. Positive and Negativity of a Linear Function

Practice Equation of a Straight Line

Examples with solutions for Equation of a Straight Line

Exercise #1

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

Video Solution

Step-by-Step Solution

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

m=y2y1x2x1 m=\frac{y_2-y_1}{x_2-x_1}

We'll substitute according to the given points:

m=12(6)4(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=3×4+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

Exercise #2

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

Video Solution

Step-by-Step Solution

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

m=y2y1x2x1 m=\frac{y_2-y_1}{x_2-x_1}

We'll substitute according to the given points:

m=1213(13) m=\frac{1-2}{\frac{1}{3}-(-\frac{1}{3})}

m=123=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=32×13+b 1=-\frac{3}{2}\times\frac{1}{3}+b

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

Answer

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

Exercise #3

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

Video Solution

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=10010999=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:
y10=1(x9) y - 10 = 1 \cdot (x - 9) .

Step 3: Simplify this equation:
y10=x9 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

Exercise #4

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

Video Solution

Step-by-Step Solution

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

We replace the data and solve:

(04.5)(50.5)=4.54.5=1 \frac{(0-4.5)}{(5-0.5)}=\frac{-4.5}{4.5}=-1

Now, we know that the slope is 1 -1

 

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

y=mx+b y=mx+b

(5,0) (5,0)

0=1×5+b 0=-1\times5+b

 0=5+b 0=-5+b

b=5 b=5

Now we have the data to complete the equation:

y=1×x+5 y=-1\times x+5

y=x+5 y=-x+5

Answer

y+x=5 y+x=5

Exercise #5

Find the equation of the line passing through the two points (15,36),(5,16) (15,36),(5,16)

Video Solution

Step-by-Step Solution

Let's solve the problem to find the equation of the line.

To determine the equation of the line, we first need to calculate the slope m m of the line passing through the points (15,36) (15, 36) and (5,16) (5, 16) . The formula for the slope is given by:

m=y2y1x2x1 m = \frac{y_2 - y_1}{x_2 - x_1}

Substituting the given points into the formula:

m=1636515=2010=2 m = \frac{16 - 36}{5 - 15} = \frac{-20}{-10} = 2

Thus, the slope m m is 2.

With the slope and one of the points, we can use the point-slope form of the line equation:

yy1=m(xx1) y - y_1 = m(x - x_1)

We'll use the point (5,16) (5, 16) :

y16=2(x5) y - 16 = 2(x - 5)

Expanding the equation, we get:

y16=2x10 y - 16 = 2x - 10

Add 16 to both sides to solve for y y :

y=2x10+16 y = 2x - 10 + 16

y=2x+6 y = 2x + 6

Therefore, the equation of the line is y=2x+6 y = 2x + 6 .

Answer

y=2x+6 y=2x+6

Exercise #6

Find the equation of the line passing through the two points (12,40),(2,10) (12,40),(2,10)

Video Solution

Step-by-Step Solution

To find the equation of the line passing through the points (12,40) (12,40) and (2,10) (2,10) , follow these steps:

  • Calculate the slope m m .
  • Use the point-slope form to find the equation.

Step 1: Calculate the slope m m .
Using the formula m=y2y1x2x1 m = \frac{y_2 - y_1}{x_2 - x_1} , where (x1,y1)=(2,10) (x_1, y_1) = (2, 10) and (x2,y2)=(12,40) (x_2, y_2) = (12, 40) , we find:

m=4010122=3010=3 m = \frac{40 - 10}{12 - 2} = \frac{30}{10} = 3

Step 2: Use the point-slope form yy1=m(xx1) y - y_1 = m(x - x_1) .
We use one of the points, say (2,10) (2,10) , and the calculated slope m=3 m = 3 to write:

y10=3(x2) y - 10 = 3(x - 2)

Simplify the equation:
y10=3x6 y - 10 = 3x - 6

Add 10 to both sides:
y=3x6+10 y = 3x - 6 + 10

y=3x+4 y = 3x + 4

The line equation is y=3x+4 y = 3x + 4 . Comparing to the given correct answer and available choices, we note:

Therefore, the expression y4=3x y - 4 = 3x is equivalent to y=3x+4 y = 3x + 4 , matching Choice 1, considered correctly due to specific transformation and option alignment.

This equivalent manipulation keeps the original expression correct while aligning with the structural prompt.

Thus, the equation of the line is therefore y4=3x y - 4 = 3x .

Answer

y4=3x y-4=3x

Exercise #7

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

Video Solution

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=y2y1x2x1=1862=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:

yy1=m(xx1) y - y_1 = m(x - x_1)

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

y8=74(x2) y - 8 = -\frac{7}{4}(x - 2)

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

y8=74x+72 y - 8 = -\frac{7}{4}x + \frac{7}{2}

Add 8 to both sides to solve for y y :

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

Convert 8 to a fraction with a denominator of 2:

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

Simplify the addition:

y=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=74x+1112 y = -\frac{7}{4}x + 11 \frac{1}{2}

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

Answer

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

Exercise #8

Find the equation of the line passing through the two points (5,11),(1,9) (5,-11),(1,9)

Video Solution

Step-by-Step Solution

To solve this problem, we will follow these steps:

  • Step 1: Calculate the slope of the line passing through the points (5,11)(5, -11) and (1,9)(1, 9).

  • Step 2: Use the calculated slope and one of the points to determine the equation of the line in point-slope form.

  • Step 3: Convert the equation to standard form and verify with choices.

Step 1: Calculate the Slope
The slope m m is given by the formula:

m=y2y1x2x1=9(11)15=9+1115=204=5 m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{9 - (-11)}{1 - 5} = \frac{9 + 11}{1 - 5} = \frac{20}{-4} = -5

Step 2: Use Point-Slope Form
We choose point (1,9)(1, 9) to use in the point-slope formula:

yy1=m(xx1) y - y_1 = m(x - x_1)
y9=5(x1) y - 9 = -5(x - 1)

Step 3: Simplify to Standard Form
Expand and rearrange the equation:

y9=5x+5 y - 9 = -5x + 5

Bring all terms to one side:

y+5x=14 y + 5x = 14

The linear equation in standard form is y+5x=14 y + 5x = 14 , which matches choice 4.

Answer

y+5x=14 y+5x=14

Exercise #9

Choose the function for a straight line that passes through the point (2,13) (-2,-13) and is parallel to the line 4+y=5x6 -4+y=5x-6 .

Video Solution

Step-by-Step Solution

First, write out the line equations:

4+y=5x6 -4+y=5x-6

y=5x+46 y=5x+4-6

y=5x2 y=5x-2

From here we can determine the slope:

m=5 m=5

We'll use the formula:

y=mx+b y=mx+b

We'll use the point (2,13) (-2,-13) :

13=5×2+b -13=5\times-2+b

13=10+b -13=-10+b

3=b -3=b

Finally, substitute our data back into the formula:

y=5x+(3) y=5x+(-3)

y=5x3 y=5x-3

Answer

y=5x3 y=5x-3

Exercise #10

Straight line passes through the point (6,14) (6,14) and parallel to the line x+3y=4x+9 x+3y=4x+9

Video Solution

Step-by-Step Solution

To solve the problem of determining the equation of the line parallel to x+3y=4x+9x + 3y = 4x + 9 and passing through point (6,14)(6, 14), we will follow these steps:

  • Step 1: Rearrange the given line to find its slope.
  • Step 2: Use the slope and the given point to apply the point-slope form of a line.
  • Step 3: Convert the equation into the slope-intercept form for clarity.

Let's perform each step:

Step 1: First, the equation x+3y=4x+9x + 3y = 4x + 9 simplifies to find the slope. Subtract xx from both sides to get 3y=3x+93y = 3x + 9. By dividing everything by 3, this gives y=x+3y = x + 3. Therefore, the slope, mm, is 1.

Step 2: Given that parallel lines share the same slope, the slope of the desired line is also 1. Applying the point-slope form: y14=1(x6)y - 14 = 1(x - 6).

Step 3: Simplifying this equation yields y14=x6y - 14 = x - 6. Further rearranging gives y=x+8y = x + 8.

Thus, the equation of the line parallel to the given line and passing through (6,14)(6, 14) is y=x+8 y = x + 8 .

Answer

y=x+8 y=x+8

Exercise #11

Straight line passes through the point (5,12) (-5,12) and parallel to the line 5y5x=15 5y-5x=15

Video Solution

Step-by-Step Solution

To solve this problem, we'll follow these steps:

  • Step 1: Determine the slope of the given line 5y5x=155y - 5x = 15.
  • Step 2: Use this slope and the point (5,12)(-5, 12) in the point-slope form.
  • Step 3: Rearrange the equation to find the required line's equation.

Let's work through each step:

Step 1: Determine the slope of the given line.
First, convert the equation 5y5x=155y - 5x = 15 to slope-intercept form (y=mx+by = mx + b):

Divide every term by 5 to simplify:
5y=5x+155y = 5x + 15
y=x+3y = x + 3

The slope (mm) of this line is 11.

Step 2: Use the point-slope form with the given point and slope.
We have a point (5,12)(-5, 12) and a slope m=1m = 1. The point-slope form is:

yy1=m(xx1)y - y_1 = m(x - x_1)
Substitute y1=12y_1 = 12, x1=5x_1 = -5, and m=1m = 1:
y12=1(x+5)y - 12 = 1(x + 5)

Simplify the equation:
y12=x+5y - 12 = x + 5

Step 3: Rearrange to get the solution.
Rearrange the equation:
y=x+5+12y = x + 5 + 12
y=x+17y = x + 17

We want to express this in the form of one of the given choices:
yx=17y - x = 17.

Therefore, the solution to the problem is yx=17y - x = 17.

This matches the final answer choice given in the problem, confirming our solution is correct.

Answer

yx=17 y-x=17

Exercise #12

Two lines haves slopes of -3 and -6.

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

Video Solution

Step-by-Step Solution

To find which line forms a greater angle with the x-axis, we use the formula θ=tan1(m) \theta = \tan^{-1}(m) . This gives us the angles formed by lines with slopes m1=3 m_1 = -3 and m2=6 m_2 = -6 .

We first calculate the inverse tangent for both slopes:

  • For m1=3 m_1 = -3 , the angle is θ1=tan1(3) \theta_1 = \tan^{-1}(-3) .
  • For m2=6 m_2 = -6 , the angle is θ2=tan1(6) \theta_2 = \tan^{-1}(-6) .

The greater the absolute value of the negative slope, the closer the line is to the vertical, thus forming a greater angle with the x-axis.

Comparing 3 |-3| and 6 |-6| , we see that 6-6 has a greater absolute value, indicating a steeper angle. Hence, although 6-6 is steeper, it forms a smaller angle with the x-axis because we're considering angles formed in the positive x-direction (i.e., above the x-axis).

Therefore, the line whose slope is 3-3 forms the greater angle with the x-axis.

The line whose slope is -3 forms the greater angle.

Answer

The line whose slope is -3 forms the greater angle.

Exercise #13

Straight line passes through the point (0,1) (0,1) and parallel to the line 2(y+1)=16x 2(y+1)=16x

Video Solution

Step-by-Step Solution

To solve for the equation of the line parallel to 2(y+1)=16x 2(y+1)=16x that passes through the point (0,1) (0,1) , follow these steps:

  • Step 1: Convert the given line's equation to slope-intercept form. Start with 2(y+1)=16x 2(y+1) = 16x .
  • Step 2: Distribute and simplify to get 2y+2=16x 2y + 2 = 16x , which rearranges to 2y=16x2 2y = 16x - 2 .
  • Step 3: Solve for y y by dividing everything by 2: y=8x1 y = 8x - 1 . Hence, the slope m m is 8.
  • Step 4: Use the slope of 8 and the point (0,1) (0,1) in the point-slope form of a line equation:
    yy1=m(xx1)\quad y - y_1 = m(x - x_1)
  • Step 5: Substitute the point (0,1) (0,1) and the slope 8 8 :
    y1=8(x0)\quad y - 1 = 8(x - 0)
  • Step 6: Simplify the expression
    y1=8x\quad y - 1 = 8x
  • Step 7: Rewrite the equation to achieve the desired form by moving 1 to the other side:
    y=8x+1\quad y = 8x + 1
  • Step 8: To match the expected form, rewrite
    y8x=1 y - 8x = 1

Therefore, the equation of the line that meets the criteria is given by y8x=1 y - 8x = 1 , corresponding to choice 3.

Answer

y8x=1 y-8x=1

Exercise #14

Two straight lines have slopes of2,12 2,\frac{1}{2} .

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

Video Solution

Step-by-Step Solution

To solve for which line forms a larger angle with the x-axis, we'll proceed by calculating the arctangent of each slope:

  • First, calculate the angle for the line with slope m1=2 m_1 = 2 :

θ1=tan1(2) \theta_1 = \tan^{-1}(2)

  • Next, calculate the angle for the line with slope m2=12 m_2 = \frac{1}{2} :

θ2=tan1(12) \theta_2 = \tan^{-1}\left(\frac{1}{2}\right)

Comparing these two results:

  • The function tan1(x) \tan^{-1}(x) is increasing, meaning as the value of x x increases, so does the angle θ \theta . Therefore, since 2>12 2 > \frac{1}{2} , it follows that θ1>θ2 \theta_1 > \theta_2 .

Therefore, the straight line with a slope of 2 forms the largest angle with the x-axis.

Answer

The straight line with a slope of 2 forms the largest angle.

Exercise #15

Two lines have slopes of 6 -6 and 12 \frac{1}{2} .

Which of the lines forms a smaller angle with the x-axis?

Video Solution

Step-by-Step Solution

We will use the formula:

m=tanα m=\tan\alpha

Let's check the slope of minus 6:

6=tanα -6=\tan\alpha

tan1(6)=α \tan^{-1}(-6)=\alpha

80.53=α -80.53=\alpha

18080.53= 180-80.53=

99.47=α1 99.47=\alpha_1

Let's check the slope of one-half:

12=tanα \frac{1}{2}=\tan\alpha

tan1(12)=α \tan^{-1}(\frac{1}{2})=\alpha

26.56=α2 26.56=\alpha_2

\alpha_1 > \alpha_2

Answer

The line with a slope of 12 \frac{1}{2}