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A line in the coordinate plane has a slope of \(4\) and a distance of \(1\) unit from the origin. Find the area of the triangle determined by the line and the coordinate axes.

 

Thank you!

 May 2, 2020
 #1
avatar+23575 
+1

Slope = rise over run    or   y /x         so   y = 4x

 

Pythag theor     x^2 + (4x)^2 = 1^2

                         17x^2 = 1

                               x^2 = 1/17

                                x = sqrt (1/17)      y = 4 sqrt (1/17)

 

Area = 1/2  b  h    =  1/2   x  y   =    1/2  sqrt 1/17   * 4 sqrt 1/17

                                                   = 2 (1/17) = 2/17    units2

 May 2, 2020
 #2
avatar+111326 
+1

See the  image below

 

 

Construct circle  x^2 + y^2  =  1

 

Construct line y = (-1/4)x

 

Sub the linear equation into the circular equation to find the x intersection of line and the circle

 

x^2  +[ (-1/4)x]^2  =  1

 

x^2  + (1/16)x^2  =  1

 

(17/16)x^2  =  1

 

x^2   =  16/17   take the  negative root

 

x=   -4/√ 17

And y  =  (-1/4)(-4/√ 17)   =  1/√ 17

 

 

The  equation  of  the line tangent to this circle  at this point will have the equation

 

y = 4 (x + 4/√ 17)  + 1/√ 17

 

The  intersection of this line  and  the x axis can be found as

 

0  =  4 ( x +4/√ 17)  + 1/√ 17

 

-1/√ 17  = 4 ( x + 4/√ 17)

 

-1/ [ 4√ 17]  = x + 4/√ 17

 

-1/[4√ 17]  - 16/[4√ 17)

 

-17/ [ 4√ 17]  =

 

-√ 17/4

 

And  the intersection of this line with the  y axis can be found as

 

y = 4[ 0 + 4/√ 17]  + 1/√ 17  =  17/√ 17  = √ 17

 

So  the area  of  this triangle  =

 

(1/2)  base  * height

 

The base length  = √ [ (√ 17)^2  + (-√ 17/4)^2 ]  =  √ [ 17  + 17/16]  =  √ [ 16 * 17 + 17] /4  =  17/4

 

So....the area of this triangle  is

 

(1/2) (17/4) (1)  =    17/8  units^2

 

 

cool cool cool

 May 2, 2020
 #3
avatar+23575 
+1

Sorry..... I gave answer for a POINT that is one unit from the origin on a line with slope = 4  

 May 2, 2020

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