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 (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 #4

Straight line passes through the point (2,13) (-2,-13) and parallel to the line 4+y=5x6 -4+y=5x-6

Video Solution

Step-by-Step Solution

First, let's arrange 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

Let's substitute the existing data back into the formula:

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

y=5x3 y=5x-3

Answer

y=5x3 y=5x-3

Exercise #5

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}

Exercise #6

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

Video Solution

Answer

y=x+1 y=x+1

Exercise #7

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

Video Solution

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 (12,40),(2,10) (12,40),(2,10)

Video Solution

Answer

y4=3x y-4=3x

Exercise #9

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

Video Solution

Answer

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

Exercise #10

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

Video Solution

Answer

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

Exercise #11

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

Answer

y=x+8 y=x+8

Exercise #12

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

Video Solution

Answer

yx=17 y-x=17

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

Answer

y8x=1 y-8x=1

Exercise #14

Two lines haves slopes of -3 and -6.

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

Video Solution

Answer

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

Exercise #15

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

Answer

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

Topics learned in later sections

  1. Graphs of Direct Proportionality Functions
  2. Representation of Phenomena Using Linear Functions