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.

Guest May 7, 2019

#1**+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

CPhill May 7, 2019