y=-x+5

y=2x-7

y=x+ 1

For the first and second straight lines, three straight lines intersect in pairs:

Simultaneous equations y=-x+5=2x-7, x=4, y= 1, so the intersection point is (4,1);

For the intersection of the second line and the third line, the same principle is (2,3);

The same is true of (8, 9) for the intersection of the first line and the third line.

For the area of triangle (see figure), S△ABC= trapezoidal adec-s △ Abd-s △ AEC =1/2× (BD+ce)? ×DE- 1/2×BD×AD- 1/2×AE×CE = 1/2×( 1+8)？ ×6- 1/2× 1×2- 1/2×4×8= 10

2. According to the meaning of the question, if two straight lines are parallel, the slopes are equal, that is, k+ 1=2, k= 1.

Let's move up by K units, and you can get.

Y=-3x+k crosses the point (2,0) and brings in: 0 =-3x2+k, so k=6.

4. A point where two straight lines intersect on the X axis. Because they are on the x axis, point y=0.

Ax+4=x+a=0, x=-a=-4/a, so a = 2.

5. If the straight line intersects the Y axis, then x=0, and the point brought in is (0, 1). Similarly, the intersection of a straight line and the X axis is (1/3,0). So the triangle area =1/2×/kloc-0 /×1/3 =1/6.

To find the distance from the origin to the straight line, we must first know the equation of the straight line perpendicular to the origin, because the slope of the straight line perpendicular to the straight line is 1/3 (defined in the book), so the straight line is y= 1/3x, and the intersection of this straight line and the straight line in the topic is (3/ 10,1/.

6. For a triangle surrounded by two straight lines and the X axis, then the two vertices of the triangle are on the X axis, and the intersection point is y=0. Find the intersection of two straight lines and the X axis, and the straight line y=x-2 intersects with the X axis at (2,0); Y=-x- 1 and the intersection with the x axis is (-1, 0); The intersection of two straight lines is (1/2, -3/2), so the base of the triangle is 2-(- 1)=3, and the height is the absolute value of -3/2, that is, 3/2.

Its area is 1/2×3×3/2=9/4.