Assume m>sqrt of 3 . Express using m the area of the triangle bounded by the lines y=(sqrt of 3)x+1,y=(negative sqrt of 3)x+1,and y=m(x-4)

Answer 1

First off with this problem you have to make sure to write the equations properly. It looks like the first 2 equations are written with the sqrt(3) and the -sqrt(3) is just multiplied to the x and no the +1 term. Also for the 3rd equation the m is multiplied to both the x and the -4 (distribution multiplication)

When x = 0 both the 1st two equation have the same value y= 1

So that is the point of intersection between those equations at point (0,1) ------> set = [*]

on the basic x,y graph (where y is in respect to the x)

now we need to find m and not given that m > sqrt(3)

The third equation is y= m(x-4) so I want to solve for m and we can also use inequality rules in comparison with m vs sqrt(3)

So if we put a sqrt(3) in the third equation subbing for m you get (sqrt(3))(x-4)

and

(sqrt(3))(x-4) < m(x-4) from the given observation that m > sqrt(3)

We have to find where the line y = m(x-4) intersects with the first function and the second function ( which are the 1st and 2nd equation respectively)

set m(x-4) = ((sqrt(3))(x)) + 1 and solve for x

note: m> sqrt(3)

so x = (1+4(m))/(m-sqrt(3)) which is actually just a number but very complicated because of the m and x variables present ( and we cannot solve for me because all we know is that m > sqrt(3) so it has many values!!

so point of intersection for 3rd and 1st equation is (((1+4m)/(m-sqrt(3))), f(((1+4m)/(m-sqrt(3))))) = [**]

and do the same with the 2nd and 3rd equation to get that x coordinate of intersection point and respective function f(x) = y coordinate of that particular intersection point! to get point:

(((1+4m)/(m+sqrt(3))), f(((1+4m)/(m+sqrt(3))))) = [***]

The area of a triangle is (1/2)( base)(height)

so base off the value of m>sqrt(3) the height will be the bisector line ( from midpoint of function 3 and the point of intersection between equation 1 and 2, and bisector line drawn from midpoint of equation equation 3 on the x,y graph

The base is just the distance of the 3rd equation (function) where it has ends being points of intersection with equation 1 and equation 2 respectively

so you answer will involve the [*],[**],[***] points of intersection and using geometry to find the area of the triangle of this system!!