HOW TO FIND THE SLOPE OF LINE AX + BY = C , SLOPE OF LINE

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how to find the slope of line ax+by = c


How To find the slope of ax +by = c


Given Equation is ax + by = c
Transferring the 1st term  containing ‘x’ to R H S
 by = c - ax 
Dividing by b to find value of  ‘y’
y = c/b - ax/b
y =  -ax/b  + c/b, 

Cancelling the terms which are going to be cancelled
Rewriting the equation compatible to y = mx+c
we get , y = (-a/b)x +c/b
Compare this equation with y = mx+c
The slope of the given equation " ax + by = c " is    m = -a/b
Hence  the slope of given line is -a/b.

Note :-So from this method we can say that slop of any line can be written as -(co eff of x /co eff of y)

How To find the slope of  x- Axis


As we know that ,The equation of X-axis is y=0

Rewriting this equation in standard form of  y = mx+c

y = 0.x + 0,

Comparing it with standard form to get m = 0,

⇒ The slope of x-axis is 0 (Zero)


How To find the slope of  Y-Axis


As we know that ,The equation of X-axis is x = 0

Rewriting this equation in standard form of  y = mx+c

0.y = 1.x + 0,

Comparing it with standard form to get m = 0,

⇒ The slope of y-axis is 0 (Zero)


How To find the slope of 4x +3y = 10


Given Equation is 4x + 3y = 10
Transforming the 1st term which contains ‘x’ to R H S
 3y = 10 - 4x
Dividing by 3 to find value of  ‘y’
   y = 10/3 - 4x/3
⇒  y =  - 4x/3  + 10/3
Rewriting the equation compatible to y = mx+c
we get , y = (-4/3)x +10/3
Comparing this equation with y = mx+c
The slope of the given equation is  m = -4/3
Hence  the slope of given line is -4/3



Note simply by applying the formula ,we can calculate the slope of  this line  -(co eff of x/co eff of y) = -4/3



How To find the slope of 2x -7y = -5


Given Equation is 2x - 7y = -5
Transforming the 1st term  containing ‘x’ to R H S
- 7y = -5 - 2x 
Dividing by -7 to find value of  ‘y’
-7y/(-7) = -5/(-7) - (2/-7)x ,


Cancelling  -ve sign of the num with -ve sign of den,we have
⇒  y =  2x/7  + 5/7
Comparing the above  equation  with y = mx+c
we get , y = (2/7)x +5/7,


The co eff of  'x' on the right hand side is the value of slope
The slope of the given equation is  m = 2/7

Hence  the slope of given line is 2/7

Simply by applying the formula ,we can calculate the slope of  this line  -(co eff of x/co eff of y) = -(-2)/(-7) = 2/7

How To find the slope of √2x +√5y = 5


Given Equation is √2x +√5y = 3
Transforming the 1st term  containing ‘x’ to R H S
√5y = 3 - √2x 
Dividing by √5 to find value of  ‘y’
√5y/(√5) = 3/(√5) - (√2/√5)x ,



Cancelling  -ve sign of the num with -ve sign of den,we have
⇒  y =  - (√2/√5)x +3/(√5)
Comparing the above  equation  with y = mx+c
we get , y = - (√2/√5)x +2/7,



The co eff of  'x' on the right hand side is the value of slope
The slope of the given equation is  m =  - (√2/√5)

Hence  the slope of given line is  - (√2/√5).

Thanks for devoting your precious time to this post How to find the slope of line when its equation is given,ax+by=c calculator, ax+by+c=0, meaning,ax+by+c=0 solve for y, ax+by=c given two points, ax+by=c what is c, slope formula,ax+by=c meaning,slope of a line formula,




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HOW TO INTEGRATE, INTEGRAL WITH SQUARE ROOT IN NUMERATOR

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Let us find out the  Integration of quadratic polynomial  in denominator, integral of square root of polynomial, integrals with square roots in denominator, integration of linear by quadratic, integral with square root in numerator ,integration of square root formula .

Integration of quadratic equation in denominator

Click here to Watch more  Integration Videos

Let us take some examples to integrate such type of example.

Problem



1st of all making the co eff off  x2  positive and unity if it is not and then    rewrite  x2 +4x -1 as sum/ difference 
of  (a + b )2  = a2 +b2 åœŸ2ab  
and c2  as follows :- 

Now write the remaining term to equalise  x2 + 4x -1,


Now writing the last term as square root of term and then multiplying the co eff of x2  with -ve sign which was earlier taken outside  .


Therefore (1) can be written as 
Integration of quadratic equation in denominator
Now use the formula written below 

Integration of quadratic equation in denominator
Here in this case 'x' will be replaced by 'x + 2' and 'a' will be replaced by √5

Where 'c' is called constant of integration.

This video will explains very easy and short method of such types of integration


Problem


HOW TO INTEGRATE INTEGRAL WITH SQUARE ROOT IN NUMERATOR

1st of all making the co eff off  x2  positive and unity if it is not and then rewrite  2x2 + x - 1 as sum / difference of 
 (a + b )2  = a2 +b2 åœŸ2ab  

and c2  as follows :- 
2x2 + x -1 =2[ x2 +(1/2) x - (1/2)  ]
2x2 + x -1 = 2[  (x )2 + 2× (1/4)(x) +(1/4)2 - (1/4)2  -1/2]

 2x2 + x -1 =  2[ {(x)2 + 2 (1/4)(x) +(1/4)2}- {(1/4)2  +1/2}]

                =  2[{(x) + (1/4)}2 - {1/16+1/2}]

                  = 2 [{(x) + (1/4)}2 - {9/16}]

                  = 2[{(x) + (1/4)}2 -  {3/4}2 ]
Now using the Result 
HOW TO INTEGRATE INTEGRAL WITH SQUARE ROOT IN NUMERATOR

Now replacing x by x + (1/4) and 'a' by 3/4 ,
The integration of this Integrand can be written as follows

Where 'c' is called constant of integration.
Now multiplying with 2 all the terms and After simplification we shall have 


In this post we discussed Integration of quadratic polynomial  in denominator,integral of   square root of polynomial,integrals with square roots in denominator, integration of linear by quadratic, integral with square root in numerator ,integration of square root formula,integration of square root formula,integral of root 2, anti derivative rules for square roots Please like it and  share this post with yours friends.

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Also Read     How to learn trigonometric identities easily
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HOW TO FIND DISTANCE BETWEEN TWO POINTS IN PLANE AND SPACE

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How to calculate the distance between 2 points in plane , distance between two points in plane , shortest distance between two points in 3d ,distance between two points in space , what is distance between two points in 3d formula , How to calculate distance in the 3d euclidean space.

HOW  TO FIND DISTANCE BETWEEN TWO POINTS IN PLANE 


Distance between two points A(x1,y1) and B(x2,y2) is given by 
HOW TOP FIND DISTANCE BETWEEN TWO POINTS IN PLANE AND SPACE

Problem 1


Suppose we have two points A(1 , 2 ) and B (3 , 4) in the plane the distance between them can be calculated as follows:-
1  . 1st of all take (x1, y1  ) as  (1 , 2 ) and (x2, y2)  as (3 , 4)  then
2.   Take the differences of x coordinates and y coordinates 
3.   Then take the square of the differences of these coordinates 
4.   After that take the sum of these squares and in the last step 
5.   Take the square root of this sum obtained in the previous step .

Then |AB| = √{(3 - 1)2 +(4 - 2)2 }
|AB| = √{(2)2 +(2)2 }
|AB| = √{4 + 4 }
|AB| = √8
|AB| = 2 √2  Units


Problem 2


Suppose we have two points A(-3 , 6 ) and B (-5 , 2) in the plane the 
distance between them can be calculated as follows:-

1  . In the 1st step  take (x1, y1  ) as  (-3 , 6 ) and (x2, y2)  as (-5 , 2)  then

2.   Take the differences of x coordinates and y coordinates 

3.   Then take the square of the differences of these coordinates 

4.   After that take the sum of these squares and in the last step 

5.   Take the square root of this sum obtained in the previous step .



Then |AB| = √{(-5 - (-3))2 +(2 - 6)2 }
 |AB| = √{(-5 + 3 )2 +(2 - 6)2 }
|AB| = √{(-2)2 +(- 4)2 }
|AB| = √{4 + 16 }
|AB| = √20
|AB| = 4√5  Units

Problem 3
Suppose we have two points A(-3 , 8 ) and B (-6 , 12) in the plane
 the distance between them can be calculated as follows:-

1 .  1st of all  take (x1, y1  ) as  (-3 , 8 ) and (x2, y2)  as (-6 , 12) 
2.   Take the differences of x coordinates and y coordinates 

3.   Then take the square of the differences of these coordinates 

4.   After that take the sum of these squares and in the last step 

5.   Take the square root of this sum obtained in the previous step .



Then |AB| = √{(-6 - (-3))2 +(12 - 8)2 }
 |AB| = √{(-6 + 3 )2 +(12 - 8 )2 }
|AB| = √{(-3)2 +(4)2 }
|AB| = √{9 + 16 }
|AB| = √25  ,  As the square root of 25 is 5 
|AB| = 5 Units

Problem 4


How to show that A(1,9) ,B(-2,0) ,C(3,15) are collinear points in the plane .
before we proceed Let us know
Collinear points :- Three or more points are said to be collinear 
points if they lies in the  same line .

We shall show these points collinear by the following formula
|AB|+|BC|=|AC| , or by using the formula that sum of distances of 1st two distances in a line is equal to the whole distance. So

|AB| = √{(-2 - 1)2 +(0 - 9)2 }
|AB| = √{(-3)2 +(0 - 9)2 }
|AB| = √{9 + 81}
|AB| = √90
|AB| = √{9×10}
|AB| = 3√10 ----------- (1)


|BC| =√{(3 - (-2))2 +(15 - 0)2 }
|BC| =√{(3 + 2))2 +(15)2 }
|BC| =√{(3 + 2))2 +(15)2 }
|BC| =√{25 +225 }
|BC| =√250
|BC| =√{25×10}
|BC| = 5√10 ----------- (2)
 ,

|AC| =√{(3 - 1)2 +(15 - 9)2 }
|AC| =√{4 + 36 }
|AC| =√40
|AC| =√{4×10}
|AC| = 2√10 ----------- (3)

Therefore from (1) , (2) and (3)

|AB| + |AC|=|BC|

3√10 +2√10 = 5√10
And also point B is common in this case, so A,B and C are collinear points 

DISTANCE BETWEEN TWO POINTS IN SPACE


how to find distance between two points

Distance between two points P(x1,y1,z1) and Q(x2,y2,z1) is given by 
HOW TOP FIND DISTANCE BETWEEN TWO POINTS IN PLANE AND SPACE





Problem 1

Suppose we have two points A(-3 , -4, 7 ) and B (-5 , 7, -9) in the plane the distance between them can be calculated as follows:-
1  . 1st take (x1, y1 ,z1,  ) as  (-3 , -3, 7 ) and (x2, y2 ,z2, )  as (-5 , 7, -9)  then
2.   Take the differences of x coordinates , y coordinates and z coordinates 
3.   Then take the square of the differences of these coordinates 
4.   After that take the sum of these squares and in the last step 
5.   Take the square root of this sum obtained in the previous step .


Then |AB| = √{(-5 - (-3))2 + (7 -  (-4))2 + (-9 - 7)2}
 |AB| = √{(-5 + 3 )2 + (7 + 4)2 + (-9 - 7)2}
 |AB| = √{(-2)2 + (11)2 + (-16)2}
|AB| = √{4 + 121 + 256 }
|AB| = √381  Units

Problem 2

Suppose we have two points A(√2 , √3, 3 ) and B (2√2 , 3√3 , 6) in the plane the distance between them can be calculated as follows:-
1  . 1st take (x1, y1 ,z1,  ) as (√2 , √3, 3 ) and (x2, y2 ,z2, )  as (2√2 , 3√3 , 6)  then
2.   Take the differences of x coordinates , y coordinates and z coordinates 
3.   Then take the square of the differences of these coordinates 
4.   After that take the sum of these squares and in the last step 
5.   Take the square root of this sum obtained in the previous step .



Then |AB| = √{(2√2  - √2 )2 + (3√3 -  √3)2 + (6 - 3)2}
 |AB| = √{(√2 )2 + (2√3)2 + (3)2}
 |AB| = √{2 + 12 + 9}
 |AB| = √23   Units

Problem 3

Show that A(-2,3,5), B(1,2,3), C(7,0,-1) are collinear points


 In order to show these points collinear . we apply the following
 formula.

|AB|+ |BC|= |AC|
|AB| = √{(1 - (-2))2 + (2 - 3)2 + (3 - 5)2}
|AB |= √{(1 + 2)2 + ( - 1)2 + (- 2)2}
|AB |= √{32 + 1 +  4 }
|AB| = √{9+ 1 +  4 }
|AB| = √14  --------------(1)

|BC| = √{(7 - 1)2 + (0 - 2)2 + (-1 - 3)2}
|BC| = √{(6)2 + ( 2)2 + (-4)2}
|BC| = √36 + 4+16}
|BC| = √36+ 4+16}
|BC| = √56
|BC| = 2√14 ----------------(2)

|AC| = √{(7 - (-2))2 + (0 - 3)2 + (-1 - 5)2}
|AC| = √{(9)2 + ( - 3)2 + (-6)2}
|AC| = √{81 + 9 + 36}
|AC| = √126
|AC| = 3√14 ----------------(3)
From (1), (2) and (3) we say that
Since |AB|+|BC|=|AC|

       √14 + 2√14 = 3√14

As B is common point , Therefore A,B and C are collinear points

Also read     Solving systems of linear equations

Conclusion 



Thanks for spending your precious time to read this post, In this post we discussed how to calculate the distance between 2 points in plane , distance between two points in plane ,distance between two points in 3d space, shortest distance between two points in 3d ,distance between two points in space , what is distance between two points in 3d formula , How to calculate distance in the 3d euclidean space.

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