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!
given 3 non collinear pts $\displaystyle (x_1,y_1), (x_2,y_2), (x_3,y_3)$, the area of the triangle formed is given by
$\displaystyle A = \frac{1}{2} \left( \left|\begin{array}{cc} x_1 & y_1 \\ x_2 & y_2 \end{array}\right| + \left|\begin{array}{cc} x_2 & y_2 \\ x_3 & y_3 \end{array}\right| + \left|\begin{array}{cc} x_3 & y_3 \\ x_1 & y_1 \end{array}\right| \right)
$
this should help you..
I don't think determinants are really necessary...
If not done, first draw a sketch.
The equation of $\displaystyle L$ is $\displaystyle y=kx$ for some real $\displaystyle k$ (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 $\displaystyle L$ and of $\displaystyle y+x=a$, that is the point $\displaystyle \left(\frac{1}{k+1}a,\frac{k}{k+1}a\right)$ (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: $\displaystyle A=\frac{1}{2}a\left|\frac{1}{k+1}a\right|$. I use absolute values because $\displaystyle k$ may (must) be negative.
Now you can easily conclude. (Solution: the equation is $\displaystyle y=-\frac{1}{2}x$ or $\displaystyle y=-\frac{3}{2}x$)
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.