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A line with slope 3 is 2 units away from the origin. Find the area of the triangle formed by this line and the coordinate axes.

 May 7, 2019
 #1
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+1

Here's one way of solving this.....

 

A perpendicular line will have a slope of (-1/3)

And the equation of this line will be   y  = (-1/3)x      (1)

 

And we can construct a circle with a radius of 2   centered at the origin.....the equation of this circle will be

x^2 + y^2  = 4     (2)

 

Sub(1) into (2)  and we have that

 

x^2 + (-1/3x)^2  = 4

x^2 + 1/9 x^2  = 4

(10/9)x^2 = 4

x^2  = 36/10

x = 6/√10           and y  =  (-1/3)(6/√10) =  -(2/√10)

 

So....the point  (6/√10, -2/√10)   will lie on the line with a slope of 3

 

And the equation of this line is

 

y = 3(x - 6/√10) - 2/√10

y = 3x - 20/√10

 

The y intercept of this line    (0, -20/√10)

The x intercept of this line  =  (20 / [ 3√10]

 

And the area of the triangle formed  = (1/2) [20/√10 ] [ 20/ [3√10 ] ]  =  400/60  =  20/3 units^2

 

Here's the graph : https://www.desmos.com/calculator/axezhbz2qb

 

 

cool cool cool

 May 7, 2019
 #2
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+1

Ty!!

Guest May 8, 2019

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