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10.3: Find the Equation of a Line


Learning Objectives

By the end of this section, you will be able to:

  • Find an equation of the line given the slope and y-intercept
  • Find an equation of the line given the slope and a point
  • Find an equation of the line given two points
  • Find an equation of a line parallel to a given line
  • Find an equation of a line perpendicular to a given line

Note

Before you get started, take this readiness quiz.

  1. Solve: (frac{2}{3} = frac{x}{5}).
    If you missed this problem, review Exercise 2.2.4.
  2. Simplify: (−frac{2}{5}(x−15)).
    If you missed this problem, review Exercise 1.10.34.

How do online retailers know that ‘you may also like’ a particular item based on something you just ordered? How can economists know how a rise in the minimum wage will affect the unemployment rate? How do medical researchers create drugs to target cancer cells? How can traffic engineers predict the effect on your commuting time of an increase or decrease in gas prices? It’s all mathematics.

You are at an exciting point in your mathematical journey as the mathematics you are studying has interesting applications in the real world.

The physical sciences, social sciences, and the business world are full of situations that can be modeled with linear equations relating two variables. Data is collected and graphed. If the data points appear to form a straight line, an equation of that line can be used to predict the value of one variable based on the value of the other variable.

To create a mathematical model of a linear relation between two variables, we must be able to find the equation of the line. In this section we will look at several ways to write the equation of a line. The specific method we use will be determined by what information we are given.

Find an Equation of the Line Given the Slope and y-Intercept

We can easily determine the slope and intercept of a line if the equation was written in slope–intercept form, y=mx+b. Now, we will do the reverse—we will start with the slope and y-intercept and use them to find the equation of the line.

Exercise (PageIndex{1})

Find an equation of a line with slope −7 and y-intercept (0,−1).

Answer

Since we are given the slope and y-intercept of the line, we can substitute the needed values into the slope–intercept form, y=mx+b.

Name the slope.
Name the y-intercept.
Substitute the values into y=mx+b.

Exercise (PageIndex{2})

Find an equation of a line with slope (frac{2}{5}) and y-intercept (0,4).

Answer

(y = frac{2}{5}x + 4)

Exercise (PageIndex{3})

Find an equation of a line with slope −1 and y-intercept (0,−3).

Answer

(y=−x−3)

Sometimes, the slope and intercept need to be determined from the graph.

Exercise (PageIndex{4})

Find the equation of the line shown.

Answer

We need to find the slope and y-intercept of the line from the graph so we can substitute the needed values into the slope–intercept form, y=mx+by=mx+b.

To find the slope, we choose two points on the graph.

The y-intercept is (0,−4) and the graph passes through (3,−2).

Find the slope by counting the rise and run.
Find the y-intercept.
Substitute the values into y=mx+b.

Exercise (PageIndex{5})

Find the equation of the line shown in the graph.

Answer

(y=frac{3}{5}x+1)

Exercise (PageIndex{6})

Find the equation of the line shown in the graph.

Answer

(y=frac{4}{3}x−5)

Find an Equation of the Line Given the Slope and a Point

Finding an equation of a line using the slope–intercept form of the equation works well when you are given the slope and y-intercept or when you read them off a graph. But what happens when you have another point instead of the y-intercept?

We are going to use the slope formula to derive another form of an equation of the line. Suppose we have a line that has slope mm and that contains some specific point ((x_{1}, y_{1})) and some other point, which we will just call (x,y). We can write the slope of this line and then change it to a different form.

(egin{array} {lrll}&m &=frac{y-y_{1}}{x-x_{1}} ext{Multiply both sides of the equation by }x−x_{1}.&mleft(x-x_{1} ight) &=left(frac{y-y_{1}}{x-x_{1}} ight)left(x-x_{1} ight) ext{Simplify.}&mleft(x-x_{1} ight) &=y-y_{1} ext{Rewrite the equation with the y terms on the left.} &y-y_{1} &=mleft(x-x_{1} ight) end{array})

This format is called the point–slope form of an equation of a line.

POINT–SLOPE FORM OF AN EQUATION OF A LINE

The point–slope form of an equation of a line with slope mm and containing the point ((x_{1}, y_{1})) is

We can use the point–slope form of an equation to find an equation of a line when we are given the slope and one point. Then we will rewrite the equation in slope–intercept form. Most applications of linear equations use the the slope–intercept form.

Exercise (PageIndex{7}): Find an Equation of a Line Given the Slope and a Point

Find an equation of a line with slope (m=frac{2}{5}) that contains the point (10,3). Write the equation in slope–intercept form.

Answer




Exercise (PageIndex{8})

Find an equation of a line with slope (m=frac{5}{6}) and containing the point (6,3).

Answer

(y=frac{5}{6}x−2)

Exercise (PageIndex{9})

Find an equation of a line with slope (m=frac{2}{3}) and containing the point (9,2).

Answer

(y=frac{2}{3}x−4)

FIND AN EQUATION OF A LINE GIVEN THE SLOPE AND A POINT.

  1. Identify the slope.
  2. Identify the point.
  3. Substitute the values into the point-slope form, (y−y_{1}=m(x−x_{1})).
  4. Write the equation in slope–intercept form.

Exercise (PageIndex{10})

Find an equation of a line with slope (m=−frac{1}{3}) that contains the point (6,−4). Write the equation in slope–intercept form.

Answer

Since we are given a point and the slope of the line, we can substitute the needed values into the point–slope form, (y−y_{1}=m(x−x_{1})).

Identify the slope.
Identify the point.
Substitute the values into (y−y_{1}=m(x−x_{1})).
Simplify.
Write in slope–intercept form.

Exercise (PageIndex{11})

Find an equation of a line with slope (m=−frac{2}{5}) and containing the point (10,−5).

Answer

(y=−frac{2}{5}x−1)

Exercise (PageIndex{12})

Find an equation of a line with slope (m=−frac{3}{4}), and containing the point (4,−7).

Answer

(y=−frac{3}{4}x−4)

Exercise (PageIndex{14})

Find an equation of a horizontal line containing the point (−3,8).

Answer

y = 8

Exercise (PageIndex{15})

Find an equation of a horizontal line containing the point (−1,4).

Answer

y = 4

Find an Equation of the Line Given Two Points

When real-world data is collected, a linear model can be created from two data points. In the next example we’ll see how to find an equation of a line when just two points are given.

We have two options so far for finding an equation of a line: slope–intercept or point–slope. Since we will know two points, it will make more sense to use the point–slope form.

But then we need the slope. Can we find the slope with just two points? Yes. Then, once we have the slope, we can use it and one of the given points to find the equation.

Exercise (PageIndex{16}): Find an Equation of a Line Given Two Points

Find an equation of a line that contains the points (5,4) and (3,6). Write the equation in slope–intercept form.

Answer

Use the point (3,6) and see that you get the same equation.

Exercise (PageIndex{17})

Find an equation of a line containing the points (3,1) and (5,6).

Answer

(y=frac{5}{2}x−frac{13}{2})

Exercise (PageIndex{18})

Find an equation of a line containing the points (1,4) and (6,2).

Answer

(y=−frac{2}{5}x+frac{22}{5})

FIND AN EQUATION OF A LINE GIVEN TWO POINTS.

  1. Find the slope using the given points.
  2. Choose one point.
  3. Substitute the values into the point-slope form, (y−y_{1}=m(x−x_{1})).
  4. Write the equation in slope–intercept form.

Exercise (PageIndex{20})

Find an equation of a line containing the points (−2,−4) and (1,−3).

Answer

(y=frac{1}{3}x−frac{10}{3})

Exercise (PageIndex{21})

Find an equation of a line containing the points (−4,−3) and (1,−5).

Answer

(y=−frac{2}{5}x−frac{23}{5})

Exercise (PageIndex{22})

Find an equation of a line that contains the points (−2,4) and (−2,−3). Write the equation in slope–intercept form.

Answer

Again, the first step will be to find the slope.

(egin{array}{lrl} ext { Find the slope of the line through }(-2,4) ext { and }(-2,-3) & & & &m &=&frac{y_{2}-x_{1}}{x_{2}-x_{1}} &m &=&frac{-3-4}{-2-(-2)} &m &= &frac{-7}{0} ext { The slope is undefined. } & & &end{array})

This tells us it is a vertical line. Both of our points have an x-coordinate of −2. So our equation of the line is x=−2. Since there is no yy, we cannot write it in slope–intercept form.

You may want to sketch a graph using the two given points. Does the graph agree with our conclusion that this is a vertical line?

Exercise (PageIndex{23})

Find an equation of a line containing the points (5,1) and (5,−4).

Answer

x = 5

Exercise (PageIndex{24})

Find an equation of a line containing the points (−4,4) and (−4,3).

Answer

x=−4

We have seen that we can use either the slope–intercept form or the point–slope form to find an equation of a line. Which form we use will depend on the information we are given. This is summarized in Table (PageIndex{1}).

To Write an Equation of a Line
If given:Use:Form:
Slope and y-interceptslope–intercepty=mx+b
Slope and a pointpoint–slope(y−y_{1}=m(x−x_{1}))
Two pointspoint–slope(y−y_{1}=m(x−x_{1}))
Table (PageIndex{1})

Find an Equation of a Line Parallel to a Given Line

Suppose we need to find an equation of a line that passes through a specific point and is parallel to a given line. We can use the fact that parallel lines have the same slope. So we will have a point and the slope—just what we need to use the point–slope equation.

First let’s look at this graphically.

The graph shows the graph of y=2x−3. We want to graph a line parallel to this line and passing through the point (−2,1).

We know that parallel lines have the same slope. So the second line will have the same slope as y=2x−3. That slope is (m_{|} = 2). We’ll use the notation (m_{|}) to represent the slope of a line parallel to a line with slope m. (Notice that the subscript ∥ looks like two parallel lines.)

The second line will pass through (−2,1) and have m=2. To graph the line, we start at (−2,1) and count out the rise and run. With m=2 (or (m=frac{2}{1})), we count out the rise 2 and the run 1. We draw the line.

Do the lines appear parallel? Does the second line pass through (−2,1)?

Now, let’s see how to do this algebraically.

We can use either the slope–intercept form or the point–slope form to find an equation of a line. Here we know one point and can find the slope. So we will use the point–slope form.

Exercise (PageIndex{25}): How to Find an Equation of a Line Parallel to a Given Line

Find an equation of a line parallel to y=2x−3 that contains the point (−2,1). Write the equation in slope–intercept form.

Answer

Does this equation make sense? What is the y-intercept of the line? What is the slope?

Exercise (PageIndex{26})

Find an equation of a line parallel to the line y=3x+1 that contains the point (4,2). Write the equation in slope–intercept form.

Answer

y=3x−10

Exercise (PageIndex{27})

Find an equation of a line parallel to the line (y=frac{1}{2}x−3) that contains the point (6,4).

Answer

(y=frac{1}{2}x+1)

FIND AN EQUATION OF A LINE PARALLEL TO A GIVEN LINE.

  1. Find the slope of the given line.
  2. Find the slope of the parallel line.
  3. Identify the point.
  4. Substitute the values into the point–slope form, (y−y_{1}=m(x−x_{1})).
  5. Write the equation in slope–intercept form.

Find an Equation of a Line Perpendicular to a Given Line

Now, let’s consider perpendicular lines. Suppose we need to find a line passing through a specific point and which is perpendicular to a given line. We can use the fact that perpendicular lines have slopes that are negative reciprocals. We will again use the point–slope equation, like we did with parallel lines.

The graph shows the graph of y=2x−3. Now, we want to graph a line perpendicular to this line and passing through (−2,1).

We know that perpendicular lines have slopes that are negative reciprocals. We’ll use the notation (m_{perp}) to represent the slope of a line perpendicular to a line with slope m. (Notice that the subscript (_{perp}) looks like the right angles made by two perpendicular lines.)

[egin{array}{cl}{y=2 x-3} & { ext { perpendicular line }} {m=2} & {m_{perp}=-frac{1}{2}}end{array}]

We now know the perpendicular line will pass through (−2,1) with (m_{perp}=−frac{1}{2}).

To graph the line, we will start at (−2,1) and count out the rise −1 and the run 2. Then we draw the line.

Do the lines appear perpendicular? Does the second line pass through (−2,1)?

Now, let’s see how to do this algebraically. We can use either the slope–intercept form or the point–slope form to find an equation of a line. In this example we know one point, and can find the slope, so we will use the point–slope form.

Exercise (PageIndex{28})

Find an equation of a line perpendicular to y=2x−3 that contains the point (−2,1). Write the equation in slope–intercept form.

Answer





Exercise (PageIndex{29})

Find an equation of a line perpendicular to the line y=3x+1 that contains the point (4,2). Write the equation in slope–intercept form.

Answer

(y=−frac{1}{3}x+frac{10}{3})

Exercise (PageIndex{30})

Find an equation of a line perpendicular to the line (y=frac{1}{2}x−3) that contains the point (6,4).

Answer

y=−2x+16

FIND AN EQUATION OF A LINE PERPENDICULAR TO A GIVEN LINE.

  1. Find the slope of the given line.
  2. Find the slope of the perpendicular line.
  3. Identify the point.
  4. Substitute the values into the point–slope form, (y−y_{1}=m(x−x_{1})).
  5. Write the equation in slope–intercept form.

Exercise (PageIndex{31})

Find an equation of a line perpendicular to x=5 that contains the point (3,−2). Write the equation in slope–intercept form.

Answer

Again, since we know one point, the point–slope option seems more promising than the slope–intercept option. We need the slope to use this form, and we know the new line will be perpendicular to x=5. This line is vertical, so its perpendicular will be horizontal. This tells us the (m_{perp}=0).

(egin{array}{lrll}{ ext { Identify the point. }} &{(3}&{,}&{-2)} { ext { Identify the slope of the perpendicular line. }} & {m_{perp}}&{=}&{0} { ext { Substitute the values into } y-y_{1}=mleft(x-x_{1} ight) .} & {y-y_{1}}&{=}&{mleft(x-x_{1} ight)} {} &{y−(−2)}&{=}&{0(x−3)} { ext { Simplify. }} & {y+2}&{=}&{0} &{y}&{=}&{-2}end{array})

Sketch the graph of both lines. Do they appear to be perpendicular?

Exercise (PageIndex{32})

Find an equation of a line that is perpendicular to the line x=4 that contains the point (4,−5). Write the equation in slope–intercept form.

Answer

y=−5

Exercise (PageIndex{33})

Find an equation of a line that is perpendicular to the line x=2 that contains the point (2,−1). Write the equation in slope–intercept form.

Answer

y=−1

In Exercise (PageIndex{31}), we used the point–slope form to find the equation. We could have looked at this in a different way.

We want to find a line that is perpendicular to x=5 that contains the point (3,−2). The graph shows us the line x=5 and the point (3,−2).

We know every line perpendicular to a vertical line is horizontal, so we will sketch the horizontal line through (3,−2).

Do the lines appear perpendicular?

If we look at a few points on this horizontal line, we notice they all have y-coordinates of −2. So, the equation of the line perpendicular to the vertical line x=5 is y=−2.

Exercise (PageIndex{34})

Find an equation of a line that is perpendicular to y=−4 that contains the point (−4,2). Write the equation in slope–intercept form.

Answer

The line y=−4 is a horizontal line. Any line perpendicular to it must be vertical, in the form x=a. Since the perpendicular line is vertical and passes through (−4,2), every point on it has an x-coordinate of −4. The equation of the perpendicular line is x=−4. You may want to sketch the lines. Do they appear perpendicular?

Exercise (PageIndex{35})

Find an equation of a line that is perpendicular to the line y=1 that contains the point (−5,1). Write the equation in slope–intercept form.

Answer

x=−5

Exercise (PageIndex{36})

Find an equation of a line that is perpendicular to the line y=−5 that contains the point (−4,−5).

Answer

x=−4

Key Concepts

  • To Find an Equation of a Line Given the Slope and a Point
    1. Identify the slope.
    2. Identify the point.
    3. Substitute the values into the point-slope form, (y−y_{1}=m(x−x_{1})).
    4. Write the equation in slope-intercept form.
  • To Find an Equation of a Line Given Two Points
    1. Find the slope using the given points.
    2. Choose one point.
    3. Substitute the values into the point-slope form, (y−y_{1}=m(x−x_{1})).
    4. Write the equation in slope-intercept form.
  • To Write and Equation of a Line
    • If given slope and (y)-intercept, use slope–intercept form (y=mx+b).
    • If given slope and a point, use point–slope form (y−y_{1}=m(x−x_{1})).
    • If given two points, use point–slope form (y−y_{1}=m(x−x_{1})).
  • To Find an Equation of a Line Parallel to a Given Line
    1. Find the slope of the given line.
    2. Find the slope of the parallel line.
    3. Identify the point.
    4. Substitute the values into the point-slope form, (y−y_{1}=m(x−x_{1})).
    5. Write the equation in slope-intercept form.
  • To Find an Equation of a Line Perpendicular to a Given Line
    1. Find the slope of the given line.
    2. Find the slope of the perpendicular line.
    3. Identify the point.
    4. Substitute the values into the point-slope form, (y−y_{1}=m(x−x_{1})).
    5. Write the equation in slope-intercept form.

Glossary

point–slope form
The point–slope form of an equation of a line with slope mm and containing the point (left(x_{1}, y_{1} ight)) is (y-y_{1}=mleft(x-x_{1} ight)).

Find the Equation Of A Line Given Its Slope And A Point On The Line

We will now look at how to use the point-slope form to get the equation of a line given its slope and a point on the line.

Example :

Find the equation of a line with slope &ndash3 and passing through (&ndash2, 1).

Step 1 : Write out the Point-slope Form

Step 2 : Substitute the slope &ndash3 and the coordinates of the point (&ndash2, 1) into the point-slope form.

Step 3 : Simplify the equation

The required equation is y = &ndash3x &minus 5

Using the Slope-Intercept to get the Equation

Example :

Find the equation of a line with slope &ndash and passing through (&ndash3, 1).

Step 1: Substitute m = &ndash , x = &ndash3 and y = 1 into the equation y = mx+ c to obtain the value of c.

Step 2: Write out the equation of the line

The required equation is or 2y = &ndash3x &ndash 7

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Ex 10.3 Class 11 Maths Question 1.
Reduce the following equations into slope- intercept form and find their slopes and the y-intercepts.
(i) x + 7y = 0,
(ii) 6x + 3y-5 = 0,
(iii) y=0
Solution:
(i) We have given an equation x + 7y = 0, which can be written in the form
⇒ 7y = – x ⇒ y = x + 0 … (1)
Also, the slope intercept form is y-mx + c …(2)
On comparing (1) and (2), we get
m = , c = 0
Hence the slope is and the y-intercept = 0.

(ii) We have given an equation 6x + 3y – 5 = 0, which can be written in the form 3y = – 6x + 5
⇒ y = – 2x + …(1)
Also, the slope intercept form is y = mx + c … (2)
On comparing (1) and (2), we get
m = – 2 and c =
i.e. slope = – 2 and the y-intercept =

(iii) We have given an equation y = 0
y = 0·x + 0 … (1)
Also, the slope intercept form is y = mx + c … (2) On comparing (1) and (2), we get
m = 0, c = 0.
Hence, slope is 0 and the y-intercept is 0.

Ex 10.3 Class 11 Maths Question 2.
Reduce the following equations into intercept form and find their intercepts on the axes.
(i) 3x + 2y – 12 = 0,
(ii) 4x – 3y = 6,
(iii) 3y + 2 = 0
Solution:
(i) Given equation is 3x + 2y – 12 = 0 We have to reduce the given equation into intercept form, i.e., …(1)
Now given, 3x + 2y = 12
⇒ ⇒ …(2)
On comparing (1) and (2), we get a = 4, b = 6 Hence, the intercepts of the line are 4 and 6.

(ii) Given equation is 4x – 3y = 6
We have to reduce the given equation into intercept form, i.e., …(1)
…(2)
On comparing (1) and (2), we get
a = and b = – 2
Hence, the intercepts of the line are and -2.

(iii) Given equation is 3y + 2 = 0
We have to reduce the given equation into intercept form, i.e.,
3y = -2
⇒ y =
The above equation shows that, it is not the required equation of the intercept form as it is parallel to x-axis.
We observe that y-intercept of the line is , but there is no intercept on x-axis.

Ex 10.3 Class 11 Maths Question 3.
Reduce the following equations into normal form. Find their perpendicular distances from the origin and angle between perpendicular and the positive x-axis.
(i) x – + 8 = 0,
(ii) y-2 = 0,
(iii) x-y = 4.
Solution:
(i) Given equation is x – + 8 = 0
x – = -8
-x + = 8 … (i)
Also,

Now dividing both the sides of (1) by 2, we get

⇒ – cos 60°x + sin 60° y = 4.

⇒ cos 120° x + sin 120° y = 4
∴ x cos 120° + y sin 120° = 4 is the required equation in normal form
∵ The normal form is x coso) + y sin⍵ = p
So, ⍵ = 120° and p = 4
⍵ Distance of the line from origin is 4 and the angle between perpendicular and positive x-axis is 120°.

(ii) Given equation is y – 2 = 0
⇒ y = 2
⇒ 0 · x + l · y = 2
⇒ x cos 90° + y sin 90° = 2 is the required equation in normal form
∵ The normal form is x cos⍵ + y sin⍵ = p
So, ⍵ = 90° and p = 2
⍵ Distance of the line from origin is 2 and the angle between perpendicular and positive x-axis is 90°.

(iii) Given equation is x – y = 4 … (1)
Also

Now dividing both the sides of (1) bt , we get

is the required equation in normal form.
∵ The normal form is x cos⍵ + y sin⍵ = p
So, P = and ⍵ = 315°
∴ Distance of the line from the origin is and the angle between perpendicular and the positive x-axis is 315°.

Ex 10.3 Class 11 Maths Question 4.
Find the distance of the point (-1, 1) from the line 12 (x+ 6) = 5(y — 2).
Solution:
The equation of line is 12(x + 6) = 5(y – 2) …(i)
⇒ 12x + 72 = 5y-10
⇒ 12x – 5y + 82 = 0
∴ Distance of the point (-1, 1) from the line (i)

Ex 10.3 Class 11 Maths Question 5.
Find the points on the x-axis, whose distances from the line are 4 units.
Solution:
We have a equation of line , which can be written as
4x + 3y – 12 = 0 … (i)
Let (a, 0) be the point on x-axis whose distance from line (i) is 4 units.

Ex 10.3 Class 11 Maths Question 6.
Find the distance between parallel lines
(i) 15x+8y-34 = 0and 15x + 8y+31 =0
(ii) |(x + y) + p = 0 and |(x + y) – r = 0.
Solution:
If lines are Ax + By + Q = 0
and Ax + By + C2 = 0, then distance between

Ex 10.3 Class 11 Maths Question 7.
Find equation of the line parallel to the line 3x – 4y + 2 = 0 and passing through the point (-2, 3).
Solution:
We have given an equation of line 3x – 4y + 2 = 0
Slop of the line(i) =
Thus, slope of any line parallel to the given line (i) is and passes through (-2, 3), then its equation is

Ex 10.3 Class 11 Maths Question 8.
Find equation of the line perpendicular to the line x – 7y + 5 = 0 and having x intercept 3.
Solution:
Given equation is x – 7y + 5 = 0 … (i)
Slope of this line =
∴ Slope of any line perpendicular to the line (i) is -7 and passes through (3, 0) then
y – 0 = -7(x – 3)
[∵ Product of slope of perpendicular lines is -1]
⇒ y = -7x + 21
⇒ 7x + y – 21 = 0, is the required equation of line.

Ex 10.3 Class 11 Maths Question 9.
Find angles between the lines + y = 1 and x + = 1.
Solution:
The given equations are
+ y = 1 … (i)
x + = 1 … (ii)
Since we have to find an angle between the two lines i.e., firstly we have to find the slopes of (i) and (ii).

Ex 10.3 Class 11 Maths Question 10.
The line through the points (h, 3) and (4, 1) intersects the line 7x – 9y – 19 = 0 at right angle. Find the value of h.
Solution:
Given points are (h, 3) and (4,1).
∴ Slope of the line joining (h, 3) & (4,1)

Ex 10.3 Class 11 Maths Question 11.
Prove that the line through the point (x1 y1) and parallel to the line Ax + By + C = 0 is A(x-x1) + B(y-y1) = 0.
Solution:
Given equation of a line is Ax + By + C = 0
∴ Slope of the above line =
i.e. slope of any line parallel to given line and passing through (x1, y1) is
Then equation is (y – y2) = (x – x1)
=> B(y – y1) = -A(x – x1)
=> A(x – x1) + B(y – y1) = 0.
Hence proved.

Ex 10.3 Class 11 Maths Question 12.
Two lines passing through the point (2, 3) intersects each other at an angle of 60°. If slope of one line is 2, find equation of the other line.
Solution:
We have given a point (2, 3), through which two lines are passing and intersects at an angle of 60°.
Let m be the slope of the other line


Ex 10.3 Class 11 Maths Question 13.
Find the equation of the right bisector of the line segment joining the points (3, 4) and (-1, 2).
Solution:
suppose the given points are A and B.
Let M be the mid point of AB.

Ex 10.3 Class 11 Maths Question 14.
Find the coordinates of the foot of perpendicular from the point (-1, 3) to the line 3x – 4y – 16 = 0.
Solution:
We have, 3x – 4y – 16 = 0
Slope of the kine(i) =
Then equation of any line ⊥ from (-1, 3) to the given line(i) is

Ex 10.3 Class 11 Maths Question 15.
The perpendicular from the origin to the line y = mx + c meets it at the point (-1,2). Find the values of m and c.
Solution:
Given, the perpendicular from the origin to the line y = mx + c meets it at the point (-1, 2)
∴ 2 = m (-1) + c … (i)
⇒ c – m = 2

Ex 10.3 Class 11 Maths Question 16.
If p and q are the lengths of perpendiculars from the origin to the lines x cosθ – y sinθ = k cos 2θ and x secθ + y cosecθ = k, respectively, prove that p 2 + 4q 2 = k 2 .
Solution:
Given p and q are the lengths of perpendiculars from the origin to the lines x cos θ – ysinθ=k cos 2θ and xsecθ+y cosec θ = k.

Ex 10.3 Class 11 Maths Question 17.
In the triangle ABC with vertices A(2, 3), 8(4, -1) and C( 1, 2), find the equation and length of altitude from the vertex A.
Solution:
We have given a AABC with the vertices, A (2, 3), B (4, -1) and C (1, 2)

Ex 10.3 Class 11 Maths Question 18.
If p is the length of perpendicular from the origin to the line whose intercepts on the axes are a and b, then show that .
Solution:
Given, p be the length of perpendicular from the origin to the line whose intercepts

We hope the NCERT Solutions for Class 11 Maths Chapter 10 Straight Lines Ex 10.3, help you. If you have any query regarding NCERT Solutions for Class 11 Maths Chapter 10 Straight Lines Ex 10.3, drop a comment below and we will get back to you at the earliest.


Determine the equation of the straight line passing through the points:

The gradient–point form of the straight line equation (EMBGB)

We derive the gradient–point form of the straight line equation using the definition of gradient and the two-point form of a straight line equation [frac = frac]

Substitute (m = dfrac) on the right-hand side of the equation [frac = m]

Multiply both sides of the equation by ((x-x_1)) [y-y_1 = m(x-x_1)]

To use this equation, we need to know the gradient of the line and the coordinates of one point on the line.


What is the equation for an undefined slope?

1 Answer. If the slope of a line is undefined, then the line is a vertical line, so it cannot be written in slope-intercept form, but it can be written in the form: x=a , where a is a constant. If the line has an undefined slope and passes through the point (2,3) , then the equation of the line is x=2 .

Also, what is an undefined slope? An undefined slope (or an infinitely large slope) is the slope of a vertical line! The x-coordinate never changes no matter what the y-coordinate is! There is no run! In this tutorial, learn about the meaning of undefined slope.

Subsequently, one may also ask, what is the equation for a zero slope?

A zero slope line is a straight, perfectly flat line running along the horizontal axis of a Cartesian plane. The equation for a zero slope line is one where the X value may vary but the Y value will always be constant. An equation for a zero slope line will be y = b, where the line's slope is 0 (m = 0).

Is a slope of 0 undefined?

The "slope" of a vertical line. A vertical line has undefined slope because all points on the line have the same x-coordinate. As a result the formula used for slope has a denominator of 0, which makes the slope undefined..


How to Find the Equation of a Line from Two Points

The following video will teach how to find the equation of a line, given any two points on that line.

Steps to find the equation of a line from two points:

  1. Find the slope using the slope formula
    • (< ext>=< ext>=frac< ext>< ext>=frac<< ext>_2-< ext>_1><< ext>_2-< ext>_1>)
    • ( ext_<1>=( ext_<1>, ext_<1>))
    • ( ext_<2>=( ext_<2>, ext_<2>))
  2. Use the slope and one of the points to solve for the y-intercept (b).
    • One of your points can replace the x and y, and the slope you just calculated replaces the m of your equation y = mx + b. Then b is the only variable left. Use the tools you know for solving for a variable to solve for b.
  3. Once you know the value for m and the value for b, you can plug these into the slope-intercept form of a line (y = mx + b) to get the equation for the line.

Additional Resources

For each of the following problems, find the equation of the line that passes through the following two points:

  1. (left ( -5,10 ight )) and (left ( -3,4 ight ))
  2. (left ( -5,-26 ight )) and (left ( -2,-8 ight ))
  3. (left ( -4,-22 ight )) and (left ( -6,-34 ight ))
  4. (left ( 3,1 ight )) and (left ( -6,-2 ight ))
  5. (left ( 4,-6 ight )) and (left ( 6,3 ight ))
  6. (left ( 5,5 ight )) and (left ( 3,2 ight ))

Solutions

Step 1: Find the slope using the formula:

We have two points,(left ( -5,10 ight )) and (left ( -3,4 ight )). We will choose (left ( -5>, 10> ight )) as point one and (left ( -3>, 4> ight )) as point two . (It does not matter which is point one and which is point two as long as we stay consistent throughout our calculations.) Now we can plug the points into our formula for slope:

The slope of the line is ( -3>), so the m in y=mx+b is ( -3>).

Step 2: Use the slope and one of the points to find the y-intercept b:

It doesn’t matter which point we use. They will both give us the same value for b since they are on the same line. We choose the point (left ( -3>, 4> ight )). Now we will plug the slope, ( -3>), and the point into y=mx+b to get the equation of the line:

Then subtract 9 from both sides:

Step 3: Plug the slope (m=( -3>)), and the y-intercept (b= ( -5>)), into y=mx+b:

Step 1: Find the slope using the formula:

We have two points, (left ( -4,-22 ight )) and (left ( -6,-34 ight )). We will choose (left ( -4>, -22> ight )) as point one and (left ( -6>, -34> ight )) as point two . (It does not matter which is point one and which is point two as long as we stay consistent throughout our calculations.) Now we can plug the points into our formula for slope:

So the slope of the line is 6 .

Step 2: Use the slope and one of the points to find b.

It doesn’t matter which point we use. They will both give us the same value for b since they are on the same line. We choose the point (left ( -4>, -22> ight )). Now we will plug the slope, 6 , and the point into y=mx+b to get the equation of the line.

Then add 24 to both sides.

Step 3: Plug the slope m = 6 , and the y-intercept b = 2 , into y=mx+b.


NCERT Solutions for Class 11 Maths Chapter 10 Straight Lines Ex 10.3

NCERT Solutions for Class 11 Maths Chapter 10 Straight Lines Ex 10.3

NCERT Solutions for Class 11 Maths Chapter 10 Straight Lines Ex 10.3 are part of NCERT Solutions for Class 11 Maths. Here we have given Class 11 Maths NCERT Solutions Straight Lines Ch 10 Exercise 10.3.

Ex 10.3 Class 11 Maths Question-1

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Ex 10.3 Class 11 Maths Question-2

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Ex 10.3 Class 11 Maths Question-3

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Ex 10.3 Class 11 Maths Question-4

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Ex 10.3 Class 11 Maths Question-5

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Ex 10.3 Class 11 Maths Question-6

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Ex 10.3 Class 11 Maths Question-7

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Ex 10.3 Class 11 Maths Question-8

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Ex 10.3 Class 11 Maths Question-9

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Ex 10.3 Class 11 Maths Question-10

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Ex 10.3 Class 11 Maths Question-11

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Ex 10.3 Class 11 Maths Question-12

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Ex 10.3 Class 11 Maths Question-13

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Ex 10.3 Class 11 Maths Question-14

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Ex 10.3 Class 11 Maths Question-15

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Ex 10.3 Class 11 Maths Question-16

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Ex 10.3 Class 11 Maths Question-17

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Ex 10.3 Class 11 Maths Question-18

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NCERT Solutions for Class 11 Maths Chapter 10 Straight Lines (सरल रेखाएँ) Hindi Medium Ex 10.3






















Maths NCERT Solutions Class 11 Maths Chapter 10 Exercise.10.3

Q1. Reduce the following equations into slope-intercept form and find their slopes and the y intercepts.

(ii) 6x + 3y – 6 = 0

Q2. Reduce the following equations into intercept form and find their intercepts on the axes.

(i) 3x + 2y – 14 = 0

(ii) 4x – 3y = 6

(iii) 3y + 2 = 0

Q3. Reduce the following equations into normal form. Find their perpendicular distances from the origin and angle between perpendicular and the positive x-axis.

(i) x – √3 y + 8 = 0

(iii) x – y = 4

Q4. Find the distance of the point (–1, 1) from the line 12(x + 6) = 5(y – 2).

Q6. Find the distance between parallel lines

(i) 15x + 8y – 34 = 0 and 15x + 8y + 31 = 0

(ii) l (x + y) + p = 0 and l (x + y) – r = 0

Q7. Find equation of the line parallel to the line 3x – 4y + 2 = 0 and passing through the point (–2, 3).

Q8. Give equation of the line perpendicular to the line x – 7y + 5 = 0 and having x intercept 3.

Q9. Calculate angles between the lines √3 x + y = 1 and x + √3 y = 1

Q10. A line passes through points (k, 3)(4, 1) intersects the line 7x – 9y – 19 = 0, at right angle. Find the value of k.

Q11. Prove that the line through the point (xa, ya) and parallel to the line Ax + By + C = 0 is A(x – xa) + B (y – ya) = 0.

Q12. The angle between the two lines is 60° at intersection and passes through the point (2, 3). Obtain the slope of a second line when the slope of first line is 2.

Q13 A line segment joining the points (4, 5) and (– 2, 3). Obtain the equation of the perpendicular bisector of the line segment.

Q14: Obtain the coordinates of the foot of perpendicular from the point (– 2, 4) to the line 3x – 4y – 16 = 0.

Q15: The normal meets point (– 2, 3), is drawn from the origin to the equation of line y = m x + c.Obtain the values of m and c.

Q16: Suppose r and s are the lengths from the lines x cos θ – y sin θ = n cos 2θ and x sec θ + y cosec θ = n to the origin perpendiculars, respectively, prove that r 2 + 4 s 2 = n 2

Q17: The vertices of the triangle PQR are P (3, 4), Q (5, – 2) and R (2, 3), obtain how long the altitude is from the vertex P and also obtain the equation.

Q 18: Suppose ‘r’ is the length of the origin to the line from perpendicular from the normal. The line has axes i and j axes as intercepts of the line, then prove that:

We hope the NCERT Solutions for Class 11 Maths Chapter 10 Straight Lines Ex 10.3, help you. If you have any query regarding NCERT Solutions for Class 11 Maths Chapter 10 Straight Lines Exercise 10.3, drop a comment below and we will get back to you at the earliest.


You have two options for writing the equation of a line: point-slope form and slope-intercept form.

I create online courses to help you rock your math class. Read more.

Both of them require that you know at least two of the following pieces of information about the line:

The . y. -intercept, . b. (the . y. -coordinate of the point at which the graph of the line crosses the . y. -axis)

If you know any two of these things, you can find the equation of the line.

Point-slope form

The equation of a line in point-slope form can be written as

In this form, . (x_1,y_1). is a point on the line, and . m. is the slope. To use this form when you know two points on the line but you don’t know the slope, first find . m. using

Then simply plug the slope . m. and the coordinates of one point . (x_1,y_1). into the point-slope form of the equation of a line.


The Significance of (K_M) and (V_)

The Michaelis-Menten model is used in a variety of biochemical situations other than enzyme-substrate interaction, including antigen-antibody binding, DNA-DNA hybridization, and protein-protein interaction. It can be used to characterize a generic biochemical reaction, in the same way that the Langmuir equation can be used to model generic adsorption of biomolecular species. When an empirical equation of this form is applied to microbial growth. The experimentally determined parameters values vary wildly between enzymes (Table (PageIndex<1>)):

Table (PageIndex<1>) : Enzyme Kinetic parameters
Enzyme (K_m) (M) (k_) (1/s) (k_/K_m) (1/M.s)
Chymotrypsin 1.5 × 10 &minus2 0.14 9.3
Pepsin 3.0 × 10 &minus4 0.50 1.7 × 10 3
Tyrosyl-tRNA synthetase 9.0 × 10 &minus4 7.6 8.4 × 10 3
Ribonuclease 7.9 × 10 &minus3 7.9 × 10 2 1.0 × 10 5
Carbonic anhydrase 2.6 × 10 &minus2 4.0 × 10 5 1.5 × 10 7
Fumarase 5.0 × 10 &minus6 8.0 × 10 2 1.6 × 10 8

While (K_m) is equal to the substrate concentration at which the enzyme converts substrates into products at half its maximal rate and hence is related to the affinity of the substrate for the enzyme. The catalytic rate (k_) is the rate of product formation when the enzyme is saturated with substrate and therefore reflects the enzyme's maximum rate. The rate of product formation is dependent on both how well the enzyme binds substrate and how fast the enzyme converts substrate into product once substrate is bound. For a kinetically perfect enzyme, every encounter between enzyme and substrate leads to product and hence the reaction velocity is only limited by the rate the enzyme encounters substrate in solution. From Equation ( ef), the catalytic efficiency of a protein can be evaluated.

This (k_/K_m) ratio is called the specificity constant measure of how efficiently an enzyme converts a substrate into product. It has a theoretical upper limit of 10 8 &ndash 10 10 /M.s enzymes working close to this, such as fumarase, are termed superefficient (Table (PageIndex<1>)).

Determining (V_m) and (K_m) from experimental data can be difficult and the most common way is to determine initial rates, (v_0), from experimental values of ([P]) or ([S]) as a function of time. Hyperbolic graphs of (v_0) vs. ([S]) can be fit or transformed as we explored with the different mathematical transformations of the hyperbolic binding equation to determine (K_d). These included:

  • nonlinear hyperbolic fit (e.g., Figure (PageIndex<1>))
  • double reciprocal plot (e.g., Lineweaver&ndashBurk plot discussed below
  • Eadie-Hofstee plot

5 Answers 5

If you start from the equation y-y1 = (y2-y1)/(x2-x1) * (x-x1) (which is the equation of the line defined by two points), through some manipulation you can get (y1-y2) * x + (x2-x1) * y + (x1-x2)*y1 + (y2-y1)*x1 = 0 , and you can recognize that:

Get the tangent by subtracting the two points (x2-x1, y2-y1) . Normalize it and rotate by 90 degrees to get the normal vector (a,b) . Take the dot product with one of the points to get the constant, c .

If you start from the equation of defining line from 2 points

you can end up with the next equation

so the coefficients will be:

My implementation of the algorithm in C

Shortcut steps: "Problem : (4,5) (3,-7)" Solve: m=-12/1 then 12x-y= 48 "NOTE:m is a slope" COPY THE NUMERATOR, AFFIX "X" Positive fraction Negative sign on between. (tip: simmilar sign = add + copy the sign) 1.Change the second set into opposite signs, 2.ADD y1 to y2 (means add or subtract them depending of the sign), 3.ADD x1 to x2 (also means add or subtract them depending of the sign), 4.Then Multiply 12 and 1 to any of the problem set. After that "BOOM" Tada!, you have your answer