Here is first.Originally Posted by nath_quam
Let the coordinates of P be (1,a) and Q be (2,b).Let O be the origin.
m is the slope.
it gives
Since B lies on CD
Hence
The unit square ABCD where A(1,0), B(1,1), C(2,1), D(2,0)
i) A line, l , passing through the origin with gradient m, cuts the sides AB and CD at P and Q respectively
is the answer to this part
i am unsure on how to approach these two parts any help would be most appreciated.
ii) For what value(s) of m does the line, l, divide the area of the square in the ratio 2:1??
iii) Another like, k, passes through the origin with gradient, n, and cuts the square through the sides AB and Bc at S and T respectively.
Show that it is not possible for k to divide the area of the square in the ratio 2:1
Thanks Nath
Join O and C.Originally Posted by nath_quam
The square is divided into two parts.
Area of the smaller part(containing the vertex B) is 1/4.
Now if the line passes through AB and BC, the triangle formed by it has area less than 1/4. But for the ratio to be 2:1, it has to be either 1/3 or 2/3(see previous post), which is not possible since 1/3>1/4 and 2/3 >1/4.
Keep Smiling
Malay
Show that it is not possible for k to divide the area of the square in the ratio 2:1
Hi:
If the line passes through side(BC), then its slope must be greater than 1/2. But, as Malay has demonstrated, the slope must have value 2/9 or 4/9, neither of which is greater 1/2.
Regards,
Rich B.