91Ó°ÊÓ

To understand the problem better, start by drawing the region \(R\) that is bounded by the x-axis, y-axis, and the line \(y=1-x\). The region is a triangle formed by the intersections of the lines and axes.

Since the region is a triangle, its area can be found using the formula \(area= \frac{1}{2}bh\), where \(b\) is the base and \(h\) is the height of the triangle. The base of the triangle is the distance along the x-axis from the y-axis to the point where \(y=1-x\) intersects the x-axis, i.e., \(x=1\). The height of the triangle is the distance from the x-axis to the point where \(y=1-x\) intersects the y-axis, i.e., \(y=1\). Therefore, $$A=\frac{1}{2}(1)(1) = \frac{1}{2}$$.

In order to find the horizontal line that bisects the region, we need to divide the area in half. Since the area of the region is \(\frac{1}{2}\), the area of each subregion should be \(\frac{1}{4}\).

Let's call the points where the line \(y=k\) intersects \(y=1-x\) and the x-axis \(A (x_A,y_A)\) and \(B (x_B, y_B)\), respectively. Since \(y_A = k\) and \(y_B = 0\), we can write \((x_B,y_B) = (x_B, 0)\). Now, we need to find \(x_B\) such that the area of the triangle formed by the points \((0,0)\), \((x_B,0)\), and \((x_B, k)\) is equal to \(\frac{1}{4}\). This area can be calculated as \(\frac{1}{2}(k)(x_B)\). Thus, $$\frac{1}{2}(k)(x_B) = \frac{1}{4}.$$ Since \(y_A=k\) and \(y_A=1-x_A\), we also have \(k=1-x_A\). Using this reality in the above equation, we now have, $$\frac{1}{2}(1-x_A)(x_B)=\frac{1}{4}$$. As point A lies on the line \(y=1-x\), we can write \(y_A=1-x_A\). But as point A's y-coordinate is k, we can substitute k in the equation and get \(k=1-x_A\). To find the value of \(x_B\), we will substitute this \(k\)'value in the area equation and solve for \(x_B\). Therefore, $$\frac{1}{2}(k)(x_B)= \frac{1}{4}$$. $$\frac{1}{2}(1-x_A)(x_B) = \frac{1}{4}$$. Now, we want to find the line that divides the area of the region equally, so we integrate the function \(y=k\) over \(x\) in the range \(0\) to \(x_B\), which will be equal to a quarter of the area of the region. The equation is: $$\int_0^{x_B} k\,dx = \frac{1}{4}$$. Since \(k=1-x_A\), we have: $$\int_0^{x_B}(1-x_A)\,dx = \frac{1}{4}$$. To find the intersection point \(A(x_A, k)\), we realize that it lies on the line \(y = 1 - x\), so \(k = 1 - x_A\). We can now rewrite the integral equation as: $$\int_0^{x_B}(1-(1-k))\,dx = \frac{1}{4}$$. $$\int_0^{x_B}k\,dx = \frac{1}{4}$$. Now integrating and solving for k, we get $$kx\Big|_0^{x_B}=\frac{1}{4}$$$$\implies k\cdot x_B=\frac{1}{4}$$. So we found that the line y=k divides the region \(R\) into two equal parts.