Coordinate representation of vectors Two straight lines are respectively x and y, intersect with point O in the same plane,∠xOy=60°, and any point P in the plane is defined as follows with respect to this coordinate system: The two lines are respectively x and y, intersect with point O in the same plane,∠xOy=60°, and any point P in the plane is defined in this coordinate system as follows: if the vector of OP=xe1+ye2(where e1 is the unit vector in the same direction as the x axis and y axis respectively), then the oblique coordinate of point P is (x, y) (1) If the oblique coordinate of point P is (1,-2), find the distance from P to O. (2) If a circle with O as its center and 1 as its radius is an equation in the oblique coordinate system xOy

Coordinate representation of vectors Two straight lines are respectively x and y, intersect with point O in the same plane,∠xOy=60°, and any point P in the plane is defined as follows with respect to this coordinate system: The two lines are respectively x and y, intersect with point O in the same plane,∠xOy=60°, and any point P in the plane is defined in this coordinate system as follows: if the vector of OP=xe1+ye2(where e1 is the unit vector in the same direction as the x axis and y axis respectively), then the oblique coordinate of point P is (x, y) (1) If the oblique coordinate of point P is (1,-2), find the distance from P to O. (2) If a circle with O as its center and 1 as its radius is an equation in the oblique coordinate system xOy

(1) P (1,-2)|OP |=√(4-1)=√3(OP perpendicular to x-axis)
(2) If P is a point on the circle: OP^2=1
(Xe1+ye2)^2=1
X^2+2xy (e1*e2)+y^2=1e1*e2=cos60=1/2
X^2+xy+y^2=1 is the round equation.

How can the concept of three points A (x1, y1), B (x2, y2), C (x3, y3) be deduced if and only if (x2-x1)(y3-y1)-(x3-x1)(y2-y1)=0? In detail, How does the concept of three points A (x1, y1), B (x2, y2), C (x3, y3) be derived if and only if (x2-x1)(y3-y1)-(x3-x1)(y2-y1)=0? Ask for detailed explanation, How to deduce the concept of three points A (x1, y1), B (x2, y2), C (x3, y3) being collinear if and only if (x2-x1)(y3-y1)-(x3-x1)(y2-y1)=0? In detail,

Three points do not coincide
That is, vectors AB and AC are collinear
Vector AB=(x2-x1, y2-y1)
Vector AC=(x3-x1, y3-y1)
The necessary and sufficient condition of collinearity of two vectors is (y3-y1)(x2-x1)=(y2-y1)(x3-x1)

In an image with an inverse ratio y=-1/x, there are three relationships (x1, y1),(x2, y2),(x3.y3), if x1> x2>0> x3,y1,y2,y3, In an image with an inverse ratio of y=-1/x, there are three relationships (x1, y1),(x2, y2),(x3.y3), if x1> x2>0> x3,y1,y2,y3,

Y3> y1> y2
Draw the image in the coordinate system and you'll see

Y3> y1> y2
Draw the image in the coordinate system and you will see

A (x1, y1), B (x2, y2), C (x3, y3) are three points on the unit circle and x1+x2+x3=0 y1+y2+y3=0 Verify x12+x22+x32=y12+y22+y32=3/2 I know the answer, but why does the outer heart of the triangular ABC coincide with the inner heart? I don't understand this.

Inside is the intersection of the angular bisectors, equidistant from the three sides.
Let M (X, Y) have aMA+bMB+cMC=0(three vectors)
MA=(X1-X, Y1-Y)
MB=(X2-X, Y2-Y)
MC=(X3-X, Y3-Y)
Then: a (X1-X)+b (X2-X)+c (X3-X)=0, a (Y1-Y)+b (Y2-Y)+c (Y3-Y)=0
X=(aX1+bX2+cX3)/(a+b+c), Y=(aY1+bY2+cY3)/(a+b+c)
M ((aX1+bX2+cX3)/(a+b+c),(aY1+bY2+cY3)/(a+b+c))

Subject M=(0,0)

The inner and outer centers coincide, and the triangle is an equilateral triangle.

The following supporting drawings:

Let A (x1, y1), B (x2, y2), C (x3, y3) be three points. (Expressed in determinant)

| X1 x2 x3|
| Y1 y2 y3|=0
|1 1 1 |

There are three points A (x1, y1), B (x2, y2), C (x3, y3)(and x1) on the image of the known inverse scale function y=x/-2009

The image is divided in the 2nd and 4th quadrants, and the drawing shows that:
Y2> y1> y3