Given two points Length PQ = (x2 - x1)2 + (y2 - y1)2 is the formula for P(x1, y1) and Q(x2, y2).

__Section Formula:__

Given two points A(x1, y1) and B(x2, y2), the coordinates of the point that splits the line segment AB in the ratio m:isare [(nx1 + mx2)/(m+n), (ny1 + my2)/(m+n)]. (To cause internal conflict)

M/n or :1 can also be used to represent the ratio m: n. Any point on the line connecting points A and B will therefore be P ((x2+x1)/(+1), (y2+y1)/(+1)).

The median AD is divided by the centroid G of the triangle ABC in a 2:1 ratio.

The coordinates of the centre are given by x=ax1+bx2+cx3, and y=(ay1+by2+cy3)/(a+b+c), respectively, if the triangle ABC's vertices are A(x1, y1), B(x2, y2), and C(x3, y3).

The intersection of two exterior and one interior angle bisectors defines the trianglecentertre. Consequently, a triangle has thrcentersres, one of which is opposite each vertex.

Think about the triangle ABC, which has the vertices A(x1, y1), B(x2, y2), and C. (x3, y3). I1, I2, and I3 are the respective centers of the ex-circles opposite of vertices A, B, and C, as shown in the figure below.

__The coordinates of I1, I2 and I3 are given by__

I1(x, y) = (–ax1+bx2+cx3)/ (a+b+c) , (–ay1+by2+cy3) / (–a+b+c)

I2(x, y) = (ax1–bx2+cx3/a–b+c, ay1–by2+cy3/a–b+c)

I3(x, y) = (ax1+bx2–cx3/a+b–c, ay1+by2–cy3/a+b–c)

Let (x1, y1), (x2, y2), and (x3, y3) respectively be the coordinates of the vertices A, B, and C of a triangle ABC. Then the area of triangle ABC is given by the modulus of

= 1/2 |[x1(y2 – y3) + x2(y3 + y1) + x3(y1 – y2)]|

__Area of an n-sided polygon__

1. Plot the points and verify that they are in the proper order.

2. Assume that the polygon's vertices are A1(x1, y1), A2(x2, y2),.., and An(xn, yn) when viewed in an anticlockwise direction.

3. The polygon's surface area is halved.

__Area of a triangle can also be expressed as__

When the triangle's vertices are taken in an anticlockwise direction, the area is positive; when taken in a clockwise direction, it is negative.

The three collinear points P1, P2, and P3 form a triangle whose area is zero, meaning that their determinant must disappear.

m = tan, where (0 180o) is the slope of the line and is the angle at which a straight line is inclined to the positive direction of the x-axis.

__Equations of a straight line in various forms:__

1.** Slope Intercept derived form:** y = mx + c, where m is the line's slope and c is its intercept

2. ** Intercept Form:** x/a + y/b = 1, where x intercept = a and y-intercept = b

3. ** Slope point form:** y – y1 = m(x – x1), where (x1, y1) is a point on the straight line and m is the slope.

** 4. Two points form:** In a coordinate plane, imagine two points A(x1, y1) and B(x2, y2) m = tan = (y-y1)/(x-x1) = (y2 - y1)/ if any point P(x, y) lies on the line connecting A and b. (x2 – x1)

** 5. Parametric form:** Consider line PQ with points Q(x1, y1). Then, any point P(x, y) has coordinates of x = x1 + r cos and y = y1 + r sin.

The line's equation is then found to be x-x1/cos = y-y1/sin = r.

6. ** Normal form: **Take a look at line l in the figure below.

The intersection points of line l are A and B. As a result, the intercepts on the x and y axes are, respectively, p/cos and p/sin.

x cos/p + y sin/p = 1 x cos + y sin = p, where p is the perpendicular distance from the origin, is the equation for line l.

The location of a point that is equidistant from two lines (having an equal perpendicular distance) is known as the angle bisector.

If a point R(p, q) lies on the bisector of two lines L1: A1x + B1y + C1 and L2: A2x + B2y + C2 = 0, then the length of the perpendicular from P to both lines must be equal. As a result, the equation for the angle bisector is given as (A1x+B1y+C1)/A12 + B12 = + (A2x+B2y+C

The aforementioned equation produces two bisectors, one for an acute angle and the other for an obtuse angle.

__How to decide a particular bisector:__

1. Think about two lines A1x + B1y + C1 = 0 and A2x + B2y + C2 = 0, respectively.

2. Make the signs of the expressions A2x3 + B1y3 + C1 and A2x3 + B2y3 + C2 identical in order to find a bisector that is in the same relative position with respect to the lines as the given point S(x3, y3). (Say affirmative)

3. The bisector to this point is given by (A1x+B1y+C1)/A12 + B12 = + (A2x+B2y+C2)/A22 + B22.

4. If the signs differ, multiply one of the equations by "-1" throughout to produce the desired positive sign. The necessary bisector will then be produced by changing the equations of the lines in the above equation.

5. The obtuse angle bisector is the bisector that points in the direction of the origin if (x3, y3) (0, 0) and A2A1 + B2B1 > 0.

__How to detect whether the bisector is of the acute or obtuse angle:__

1. Find both of the bisectors first.

2. Next, determine the angle that one of them makes with the starting line.

3. The relationship between all of the path's coordinates that holds for no other points besides those that are on the path is known as the equation to the locus.

__How to find the equation of the locus of a point:__

1. Assign P the coordinates (h, k) if we are determining the equation of the locus of a point P.

2. Attempt to formulate the conditions as equations involving both known and unknowable variables.

3. Remove the parameters, leaving only h, k, and the known quantities.

4. Change h and k in the equation to x and y, respectively. The equation of the locus of p is what is produced.

The product of the slopes of the two lines is -1 if the two lines are perpendicular to one another, or m1m2 = -1.

Any line that is parallel to the equation axe + by + c = 0 has the form bx - ay + k = 0.

M1 = M2 if the two lines are coincident or parallel.

Any line that is perpendicular to axe + by + c = 0 has the form axe + by + k = 0.

Let's say there are two lines, l1 and l2, each with a different slope (m1). Thus, tan = m1 and tan = m2. Depending on the side taken into account, the angle between them is either - or - ( - ).

**The equations a1x + b1y + c1 = 0 I The formula a2x + b2y + c2 = 0 (ii) Comprise**

1. Lines that intersect if a 1/a2 > b 1/b2

2. If a1/a2 = b1/b2, parallel lines exist.

3. Lines that are coincident if a1/a2 = b1/b2 = c1/c2

__To determine the separation between the point P(x1, y1) and the line l1 with the equation x cos + y sin = p:__

1. Assume that l2 is a line that passes through P parallel to l1.

2. Define d as P's separation from l1.

3. The normal from O to l2 is then p + d in length. So, x cos + y sin = p + d is the equation for l2.

4. Because it lies on P(x1, y1), it must satisfy the equation.

"x1 cos" and "y1 sin" equal "p + d"

D is equal to x1 cos + y1 sin - p.

** Complete Distance Formula:** If the point P and the origin O are located on opposite sides of a line as shown in the following diagram, then

D is equal to -(x1 cos + y1 sin - p).

Consequently, the full distance formula is

D is equal to + (x1 cos + y1 sin - p).

Thus, d = + (ax1+by1+c)/a2+b2 is obtained.

For Detail information about Straight Lines please watch below video by Chandrasekhar Sir

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