1 pointA line 𝑙l is such that its segment between the lines 𝑥−𝑦+2=0x−y+2=0 and 𝑥+𝑦−1=0x+y−1=0 is internally bisected at the point (1,1.5)(1,1.5). What is the equation of the line 𝑙l?𝑥+2𝑦=1x+2y=1𝑥−2𝑦=3x−2y=3𝑦=3𝑥y=3x𝑥=1x=1

The given lines are x - y + 2 = 0 and x + y - 1 = 0. The point where the line l is bisected is (1, 1.5).

First, we find the intersection point of the two given lines.

To do this, we can add the two equations:

x - y + 2 + x + y - 1 = 0 + 0 2x + 1 = 0 x = -1/2

Substitute x = -1/2 into the first equation:

-1/2 - y + 2 = 0 -y + 3/2 = 0 y = 3/2

So, the intersection point of the two given lines is (-1/2, 3/2).

The midpoint formula is M = [(x1 + x2)/2 , (y1 + y2)/2].

We know that the midpoint is (1, 1.5) and one of the endpoints is (-1/2, 3/2). Let's denote the other endpoint as (x, y).

Setting up the midpoint formula gives us:

1 = (-1/2 + x)/2 1.5 = (3/2 + y)/2

Solving these equations gives x = 5/2 and y = 3/2.

The equation of a line in slope-intercept form is y = mx + b.

The slope m is (y2 - y1) / (x2 - x1) = (3/2 - 3/2) / (5/2 - -1/2) = 0.

The y-intercept b is y - mx = 3/2 - 0 = 3/2.

So, the equation of the line l is y = 3/2.