2. As shown in the figure, the coordinate of point A is known as (1, 0), and point B moves on the straight line Y =-X. When the line segment AB is the shortest, find the coordinate of point B. ..
3. As shown in the figure, in the rectangular coordinate system, the coordinate of the vertex B of the right-angle OABC is (15,6), and the straight line y= 1/3x+b just divides the right-angle OABC into two parts with equal areas, so as to find the value of B. ..
4. As shown in the figure, in the plane rectangular coordinate system, the straight line Y = 2x-6 intersects with the X axis and Y axis at points A and B respectively, and point C is on the X axis. If △ABC is an isosceles triangle, try to find the coordinates of point C. ..
5. In the plane rectangular coordinate system, A (1, 4) and B(3 1) are known, and P is a point on the coordinate axis. (1) What is the minimum value of AP+BP when the coordinates of P are? What is the maximum value of AP-BP when the coordinate of P is?
6. As shown in the figure, it is known that the image of the linear function is at point A in the second quadrant, the X axis is at point B (-6,0), the area of △AOB is 15, and AB=AO. Find the linear function and the analytical expression of the linear function.
7. It is known that the image of the linear function passes through the point (2, 20), and the area of the triangle enclosed by it and the two coordinate axes is equal to 1. Find the expression of this linear function.
8. The image of the proportional function Y=k 1x and the image of the linear function y=k2x-9 intersect at point p (3, -6).
Find the values of K 1 and K2.
If the image of linear function y=k2x-9 intersects with the X axis at point A, find the coordinates of point A..
9. The side length of a square ABCD is 4. Put this square in the plane rectangular coordinate system, so that AB is on the negative semi-axis of X axis, and the coordinate of point A is (-1, 0).
(1) The straight line Y passing through point C =-4x- 16 intersects with the X axis at point E, and the area of quadrilateral AECD is found;
(2) If the straight line L passes through the point E, the square ABCD is divided into two parts with the same area, and the analytical expression of the straight line L is found. ..
10. In the plane rectangular coordinate system, the image of the linear function y = kx+b (where b is less than 0) intersects with the X-axis, Y-axis and straight line x=4 at points A, B and C respectively, and the straight line x=4 intersects with the X-axis at point D, and the area of the quadrilateral OBCD is 10. If the abscissa of a is -65438+,
1 1. In the plane rectangular coordinate system, the image of a linear function passes through point B (-3,4) and intersects with the Y axis at point A, OA=OB: Find this linear resolution function.
12, as shown in the figure, A and B are points on the left and right sides of the origin on the X-axis, P(2, m) is in the first quadrant, the straight line PA intersects the Y-axis at the C (0, 2), and the straight line PB intersects the Y-axis at the D-axis, with SAOP=6.
Find the area of (1)△COP.
(2) Find the coordinates of point A and the value of m;
(3) If SBOP =SDOP, find the analytical formula of straight line BD.
The images of 13 and linear function y=- x+ 1 intersect the X axis and the Y axis at points A and B respectively, and an equilateral △ABC is made with AB as the side in the first quadrant.
(1) Find the area of △ABC and the coordinates of point C;
(2) If there is a point P(a,) in the second quadrant, try to express the area of quadrilateral ABPO with an algebraic expression containing a. ..
(3) Is there a point M on the X axis that makes △MAB an isosceles triangle? If it exists, please write the coordinates of point M directly; If it does not exist, please explain why.
It shows 14, and the image of the known proportional function y=k 1x and linear function y=k2x+b, and their intersection point is a (-3,4) and OB= OA.
(1) Find the analytic expressions of proportional function and linear function;
(2) Find the area and perimeter of △AOB;
(3) Is there a point P in the plane rectangular coordinate system, so that P, O, A and B are the four vertices of the right-angled trapezoid? If it exists, please write the coordinates of point P directly; If it does not exist, please explain why.
15 As shown in the figure, it is known that the image of the linear function y=x+2 intersects with the X axis at point A and with the Y axis at point C,
(1) Find the degree of ∠ Cao;
(2) If the straight line y=x+2 translates two units to the left along the X axis, try to find the analytical formula of the translated straight line;
(3) if the image of the positive proportional function y=kx (k≠0) and the image of y=x+2 intersect at point B, and ∠ ABO = 30, find the length of AB and the coordinates of point B. ..
The images of 16 and linear function y= x+2 intersect with X axis and Y axis at points A and B respectively, and make equilateral △ABC in the second quadrant with AB as the edge.
(1) Find the coordinates of point C;
(2) There is a point M(m, 1) in the second quadrant, so that S△ABM =S△ABC, and find the coordinates of this point m;
(3) Does point C (2 2,0) have a point P on the straight line AB, so that △ACP is an isosceles triangle? If it exists, find the coordinates of point P; If it does not exist, explain why.
17, it is known that the proportional function y=k 1x and the linear function y=k2x+b intersect at point a (8,6), the linear function intersects with the x axis at point b, and OB=0.6OA Find the analytical expressions of these two functions.
18. It is known that the image of linear function y=x+2 passes through point A(2, m). Intersect with x axis at point c, and find the angle AOC.
It is known that the image of 19 of function y=kx+b is parallel to the image of linear function y=x+ 1 through point a (4 4,3), and point b (2 2,m) is on the image of linear function y = kx+b.
(1) Find the expression of this linear function and the value of m?
(2) If there is a moving point P(x, 0) on the X axis, the distances to the fixed points A(4, 3) and B(2, m) are PA and PB, respectively. What is the abscissa of point P, and the value of PA+PB is the smallest?
answer
3. The shortest distance from a point to a straight line is that the point is perpendicular to the straight line. Because the angle between the straight line and X is 45 degrees, ABO is the root of an isosceles right triangle AB=BO=2 times AO = 1BO = 2.
In b, a vertical line is perpendicular to xy, and the intersection of the vertical line and the axis is the coordinate of B.
Because it is still an isosceles right triangle, according to the above * *, the coordinate of point B is (0.5, -0.5).
7. The analytical formula of linear function is y=8x+4 or y=(25/2)x-5. Let the linear function be y=kx+b, then its intersection with the two coordinate axes is (-b/k, 0)(0, b), so there are 20 = 2x+b, |-b/k×. K2=25/2，b2=-5。 Therefore, the analytical formula of the linear function is y=8x+4 or y=(25/2)x-5.
8. Because both the proportional function and the linear function pass through (3, -6).
This is a dual-function image.
Therefore, when x=3, y=-6 is substituted respectively.
k 1= -2 k2= 1
If the linear function image intersects the X axis at point A, the ordinate of A is 0.
Substitute y=0 into y=x-9 to get x=9.
So a (9 9,0)
Example 4: abscissa A =- 1/2, ordinate =0.
0=-k/2+b，k=2b
The abscissa of point C is 4, and the ordinate y is 4k+b is 9b.
The abscissa of point b =0, and the ordinate y = b.
Sobcd=(\9b\+\b\)*4/2= 10
10\b\=5
\b\= 1/2
b= 1/2，k=2b= 1 y=x+ 1/2
b=- 1/2，k=- 1 y=-x- 1/2
\b\ represents the absolute value of b.
1 1、? Solution: Let this principal resolution function be y = kx+b.
∫y = kx+b passes through point b (-3,4), and intersects with the y axis at point a, OA=OB.
∴{-3k+b=4
{3k+b=0
∴{k=-2/3
{b=2
The resolution function is y =-2/3x+2.
? Solution 2 According to Pythagorean theorem, OA=OB=5,
Therefore, it is divided into two situations:
When a (0,5), substitute b (-3,4) into y=kx+b, y=x/3+5,
When A(0, -5), substitute B(-3, 4) into y=kx+b and y=3x+5.
12, make auxiliary line PF perpendicular to y axis at point f and auxiliary line PE perpendicular to x axis at point e.
(1) Find the point cloud of S triangle.
Solution: S triangle COP =1/2 * oc * PF =1/2 * 2 = 2.
(2) Find the coordinates of point A and the value of p.
Solution: It can be proved that triangle CFP is equal to triangle COA, so there is
PF/OA = FC/OC。 Substituting PF=2 and OC=2, there is FC * OA = 4. (1 formula)
Because AOP=6 of the S triangle is 6, according to the triangle area formula, there is S = 1/2 * AO * PE = 6, so AO * PE = 12. (Formula 2).
Where PE = OC+FC = 2+FC, so the formula (2) is equal to AO * (2+FC) = 12. (3)
By solving the equations composed of (1) and (3), we can get AO = 4 and FC = 1.
p = FC + OC = 1 + 2 = 3。
So the coordinates of point A are (-4,0), the coordinates of point P are (2,3), and the value of p is 3.
(3) If S triangle BOP=S triangle DOP, find the analytical formula of straight line BD.
Solution: Because S triangle BOP=S triangle DOP, there is (1/2) * ob * PE = (1/2) * PF * OD, that is.
(1/2) * (OE+be) * PE = (1/2) * pf * (of+FD), and the value obtained above is replaced by
(1/2) * (2+be) * 3 = (1/2) * 2 * (3+FD), which means 3BE = 2FD.
And because: FD: DO = PF: OB means FD:(3+FD) = 2:(2+BE), we know that the coordinate of BE = 2. B is (4,0).
Substitute BE=2 into the above formula 3BE=2FD, and FD = 3. The d coordinate is (0,6).
Therefore, the analytical formula of straight line BD can be obtained as follows:
y = (-3/2)x + 6
17 image, the proportional function y=k 1x and the linear function y=k2x+b intersect at point a (8,6), so there is 8k 1 = 6...( 1).
8k2+b = 6...(2) and OA= 10, so OB=6 is the coordinate (6,0) of point B, so 6k2+b = 0...(3) Solution (1)(2)(3) obtains k/kloc-.
oa=√(8^2+6^2)= 10,ob=6,b(6,0),k 1=6/8=0.75
Proportional function y=0.75x, linear function y=3x- 18.
18 and the image of linear function y=x+2 pass through point a (2, m), and there are
m=2+2=4，
It intersects the X axis at point C. When y=0, x=-2.
The area of triangle aoc is:1/2 * | oc | m | =1/2 * |-2 | | 4 | = 4 square units.
19, solution: two straight lines are parallel and have equal slopes.
Therefore, k= 1, that is, the linear equation is y = x+B. After passing through point (4,3), there are:
b=- 1
Therefore, the expression of the linear function is: y=x- 1.
After passing through point (2, m), the following is replaced:
m= 1
2) A (4,3), B (2, 1) Minimizes PA+PB, then P, A and B are in a straight line.
The straight line equation of AB is:
(y-1)/(3-1) = (x-2)/(4-2) The intersection (x, 0) is replaced as follows:
(0- 1)/2=(x-2)/2
x= 1
That is, when the abscissa of point P is 1, the value of PA+PB is the smallest.