Let a be a real number. Determine the equation of a line L in the plane, passing through the origin such that the area of the figure bounded by L and the two lines:
x + y - a = 0, x = 0
is a^2.
I don't know even how to begin. Any advice? THanks!
Let a be a real number. Determine the equation of a line L in the plane, passing through the origin such that the area of the figure bounded by L and the two lines:
x + y - a = 0, x = 0
is a^2.
I don't know even how to begin. Any advice? THanks!
I don't think determinants are really necessary...
If not done, first draw a sketch.
The equation of is for some real (the case of a vertical line can be ruled out).
We are interested in the area of the triangle formed by the origin (0,0), the point (0,a) and the intersection of and of , that is the point (why ?). The area is given by half the product of the lengths of the base and the corresponding height. Choose [(0,0),(0,a)] as the base. Then you get that the area is: . I use absolute values because may (must) be negative.
Now you can easily conclude. (Solution: the equation is or )
Laurent.
I wouldn't even have thought of that as a Linear Algebra problem! It's purely Cartesian geometry.
The line x+ y= a passes through (a, 0) and (0, a). It, together with x= 0 and y= kx (any line through the origin can be written in that form) form a triangle with vertices (0, 0), (0, a) and the point where x+y= a and y= kx cross: call it (X, Y). That triangle has base of length a and height X and so area (1/2)aX and we want that equal to a^2: (1/2)aX= a^2 so X= 2a.
Now, the point where y= kx and x+ y= a cross is given by x+ kx= (1+k)x= a or x= a/(1+k). That is X= 2a= a/(1+k) so 1+ k= 1/2 and k= -1/2.
As a check, If y=(-1/2)x, then its graph intersects x+ y= a at x-(1/2)x= (1/2)x= a so x= 2a and y= -a. The triangle has vertices (0, 0), (0, a) and (2a, -a). It has area (1/2)(a)(2a)= a^2.