We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive pseudonymised information about your use of our website. cookie policy and privacy policy.
 
+0  
 
0
90
2
avatar

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
avatar+104647 
+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
avatar
+1

Ty!!

Guest May 8, 2019

12 Online Users

avatar