The line l1y = - 2x + m intersects with the line l2y = x + 1 at the point P (a, 2) to find the area of the triangle surrounded by L1, L2 and x-axis. I have found that the analytical expression of the line L1 is y = - 2x + 4

The line l1y = - 2x + m intersects with the line l2y = x + 1 at the point P (a, 2) to find the area of the triangle surrounded by L1, L2 and x-axis. I have found that the analytical expression of the line L1 is y = - 2x + 4

"I have found that the analytic expression of the straight line L1 is y = - 2x + 4"
The point P (1,2) has also been obtained
Now find out the intersection coordinates of L1, L2 and X axis as (- 1,0), (2,0)
S＝1/2 ×[2－（-1）]×（2－0）
＝1/2 ×3×2＝3

Known: line L1: y = 3x + 10, L2: y = - 2x-5, find: the area of triangle formed by two lines and Y axis

:L1y=3x+10 b1=10
L2：y=-2x-5,b2=-5
The intersection coordinates of the two lines are (- 3,1)
The area of triangle enclosed by two straight lines and Y axis is 1 / 2 (absolute value B1 + absolute value B2) * - the absolute value of 3 is 45 / 2

Given the line L1: y = x + 1, L2: y = - X-1, L3: y = - 2x + 4, the area of the triangle formed by these three lines is

It is known that L1: y = x + 1, L2: y = - X-1, L3: y = - 2x + 4
L1 and L2 intersect a (- 1,0)
Intersection of L1 and L3 with B (1,2)
The intersection of L2 and L3 with C (5, - 6)
AB = √[2²+2²] = 2√2
The distance between C and L1 is 1 / √ 2
So,
SABC = 2√2 * 1/√2 * (0.5) = 1
Hope to adopt~~~

The analytic formula of the line y = 2x-6 with respect to the y-axis symmetry is ()
A. y=2x+6B. y=-2x+6C. y=-2x-6D. y=2x-6

Two points can be found from the line y = 2x-6: (0, - 6) and (3, 0) the symmetry points of these two points about the Y axis are (0, - 6) (- 3, 0), then these two points are on the line y = 2x-6 about the Y axis symmetry y = KX + B, then B = - 6, - 3K + B = 0, and the solution is k = - 2.. therefore: C