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.
+128475 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
