Determine the equation of the line who's sum of it's intercepts on the axes is 20 and it passes through the point (1,2).

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hala718's profile pic

hala718 | High School Teacher | (Level 1) Educator Emeritus

Posted on

let the equation be :

x/a + y/b = 1     where x and b are the axis.

It is given that sum os axis = 20

==> a + b = 20

==> a= 20 - b

Substitute:

==> x/(20-b) + y/b = 1

We know tha the point (1,2) passes through th eline. Then the point (1,2) should verify the equation:

==> 1/(20-b) + 2/b = 1

==> Now multiply by b(20-b)

==> b + 2(20-b)  = b(20-b)

==> b + 40 - 2b = 20b - b^2

Group similar terms:

==> b^2 - 21b + 40 = 0

==> b1= [21+ sqrt(281)]/2 

 ==> a1= 20 - [21+sqrt281]/2= 19+ sqrt281]/2

==> b2= [ 21- sqrt(281)/2

==> a2= [ 19 - sqrt281]/2

Top Answer

giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

We'll write the intercept form of the equation of the line:

x/a + y/b = 1,

where a is x intercept and b is y intercept.

The sum of intercepts on the axis is 20.

a+b = 20 => b = 20 - a

We'll re-write the equation:

x/a + y/(20 - a) = 1

We know that the line passes through the point (1,2).

We'll substitute the coordinates of the point into the equation of the line:

1/a + 2/(20-a) = 1

We'll calculate LCD:

(20-a) + 2a = a(20-a)

We'll remove the brackets:

20 - a + 2a = 20a - a^2

We'll move all terms to one side:

20 + a - 20a + a^2 = 0

a^2 - 19a + 20 = 0

We'll apply the quadratic formula:

a1 = [19+sqrt(281)]/2

a2 = [19-sqrt(281)]/2

We'll write the equations for both values of a:

For a = [19+sqrt(281)]/2 =>

2x/[19+sqrt(281)] + 2y/[40 - 19 -sqrt(281)] = 1

2x/[19+sqrt(281)] + 2y/[21 -sqrt(281)] = 1

For [19-sqrt(281)]/2 =>  

2x/[19-sqrt(281)] + 2y/[40 - 19 +sqrt(281)] = 1

2x/[19-sqrt(281)] + 2y/[21+sqrt(281)] = 1

neela's profile pic

neela | High School Teacher | (Level 3) Valedictorian

Posted on

We can write the equation of  line in the form:

x/a+y/b = 1, where a and b are x and y intercepts of the line.

Given a+b = 20. Therefore we write b = 20-a, and rewrite the equation:

x/a+y/(2-a) = 1. Since this line passes through (1.2), it should satisfy a/a+y/(20-a) = 1. So substitute x = 1, 1nd y = 2 in this equation and solve for a.

1/a+2/(20-a) = 1. Multiply by a(20-a)

20-a+2a = (20-a)a

20+a = 20a -a^2

20+a-20a+a^2 = 0

a^2-19a+20 = 0

(a-20)(a+1) = 0

 a- 19 = 0 Or a+1 = 0.

If a= 19, then b = 20-19 = 1,

If a = -1, then b = 20- -1 = 21.

Therefore the equation of the lines are:

x/19 +y/1 = 1 Or x+19y = 19.....(1)

Also x/21+y/(-1) = 1 or -x+21y = -21 or

x-21y = 21..........(2)

(1) and (2) are the lines required

 

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