Inter 1st Year Maths 1B Locus Formulas

Use these Inter 1st Year Maths 1B Formulas PDF Chapter 1 Locus to solve questions creatively.

Intermediate 1st Year Maths 1B Locus Formulas

→ PQ = \(\sqrt{\left(x_{1}-x_{2}\right)^{2}+\left(y_{1}-y_{2}\right)^{2}}\)

→ OP = \(\sqrt{x_{1}^{2}+y_{1}^{2}}\)

→ P divides A(x1, y1) and B(x2, y2) in the ratio m : n

→ Co-ordinates of P are \(\left(\frac{m x_{2}+n x_{1}}{m+n}, \frac{m y_{2}+n y_{1}}{m+n}\right)\)

→ Midpoint = \(\left(\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2}\right)\)

→ ΔABC = \(\frac{1}{2}\)[x1 (y2 – y3) + x2 (y3 – y1) + x3(y1 – y2)] = \(\frac{1}{2}\)\(\left|\begin{array}{lll}
x_{1} & y_{1} & 1 \\
x_{2} & y_{2} & 1 \\
x_{3} & y_{3} & 1
\end{array}\right|\)

→ Area of the Quadrilateral = \(\frac{1}{2}\)|x1(y2 – y4) + x2(y3 – y1) + x3(y4 – y2) + x4(y1 – y3)|

→ Centroid G = \(\left(\frac{x_{1}+x_{2}+x_{3}}{3}, \frac{y_{1}+y_{2}+y_{3}}{3}\right)\)

→ Incentre I = \(=\left(\frac{a x_{1}+b x_{2}+c x_{3}}{a+b+c}, \frac{a y_{1}+b y_{2}+c y_{3}}{a+b+c}\right)\)

→ Ex-centre I1 = \(\left(\frac{-a x_{1}+b x_{2}+c x_{3}}{-a+b+c}, \frac{-a y_{1}+b y_{2}+c y_{3}}{-a+b+c}\right)\)

Inter 1st Year Maths 1B Locus Formulas

Distance Between two points:

  • The distance between two points A(x1, y1) and B(x2, y2)
    AB (orBA) = \(\sqrt{\left(x_{1}-x_{2}\right)^{2}+\left(y_{1}-y_{2}\right)^{2}}\)
  • The distance from origin O to the point A(x1, y1) OA = \(\sqrt{x_{1}^{2}+y_{1}^{2}}\)
  • The distance between two points A(x1, 0) and B(x2, 0) lying on the X – axis is AB= \(\sqrt{\left(x_{1}-x_{2}\right)^{2}+(0-0)^{2}}=\sqrt{\left(x_{1}-x_{2}\right)^{2}}\) = |x1 – x2|
  • The distance between two points C(0, y1) and D(0, y2) lying on the Y-axis is CD = |y1 – y2|

Section Formula:

  • The point P which divides the line segment joining the points A(x1, y1). B(x2, y2) in the ratio m : n internally is given by P = \(\frac{m x_{2}+n x_{1}}{m+n}, \frac{m y_{2}+n y_{1}}{m+n}\)
  • If P divides in the ratio m:n externally then P = \(\frac{m x_{2}-n x_{1}}{m-n}, \frac{m y_{2}-n y_{1}}{m-n}\) (m’ n)

Note: If the ratio m : n is positive then P divides internally and if the ratio is negative P divides externally.

Mid Point:
The mid point of the line segment joining A(x1, y1) and B(x2, y2) is \(\left(\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2}\right)\)

Points Of Trisection:
The points which divide the line segment \(\overline{A B}\) in the ratio 1: 2 and 2: 1 (internally) are called the points of trisection of \(\overline{A B}\)

Area Of A Triangle:
The area of the triangle formed by the points A(x1, y1), B(x2, y2) C(x3, y3) is
Area = \(\frac{1}{2}\)|(x1y2 – x2y1) +(x2y3 – x3y2) + (x3y1 – x1y3)|
i.e., Area of ABC = \(\frac{1}{2}\)|Σ(x1y2 – x2y1)|

Note:

  • The area of the triangle formed by the points (x1, y1)(x2, y2).(x3, y3) is the positive value of the determinant \(\frac{1}{2}\)\(\left|\begin{array}{ll}
    x_{1}-x_{2} & y_{1}-y_{2} \\
    x_{2}-x_{3} & y_{2}-y_{3}
    \end{array}\right|\)
  • The area of the triangle formed by the points (x1, y1)(x2, y2) and the origin is \(\frac{1}{2}\)|x1y2 – x2y1|

Area Of A Quadrilateral:
The area of the quadrilateral formed by the points (x1, y1) (x2, y2), (x3, y3), (x4, y4) taken in that order is
\(\frac{1}{2}\)|x1y2 – x2y1 + x2y3 – x3y2 + x3y4 – x4y3 + x4y1 – x1y4|

Note:
The area of the quadrilateral formed by the points (x1, y1) (x2, y2) (x3, y3),(x4, y4) taken in order is
Area = \(\frac{1}{2}\left|\begin{array}{ll}
x_{1}-x_{3} & y_{1}-y_{3} \\
x_{2}-x_{4} & y_{2}-y_{4}
\end{array}\right|\)

Inter 1st Year Maths 1B Locus Formulas

Centres Of A Triangle:
Median: In a triangle, the line segment joining a vertex and the mid point of its opposite side is called a median of the triangle. The medians of a triangle are concurrent.
The point of concurrence of the medians of a triangle is called the centroid (or) centre of gravity of the triangle. It is denoted by G.

In Centre Of A Triangle:
Internal bisector : The line which bisects the internal angle of a triangle is called an internal angle bisector of the triangle.
The point of concurrence of internal bisectors of the angles of a triangle is called the incentre of the triangle, it is denoted by I.
I = \(\left(\frac{a x_{1}+b x_{2}+c x_{3}}{a+b+c}, \frac{a y_{1}+b y_{2}+c y_{3}}{a+b+c}\right)\)

Ex-centres Of A Triangle:
The point of concurrence of internal bisector of angle A and external bisectors of angles B, C of ABC is called the ex-centre opposite to vertex A. It is denoted by I1. The excentres of ABC opposite to the vertices B, C are respectively denoted by I2, I3.

  • I1 = Excentre opposite to A = \(\left(\frac{-a x_{1}+b x_{2}+c x_{3}}{-a+b+c}, \frac{-a y_{1}+b y_{2}+c y_{3}}{-a+b+c}\right)\)
  • I2 = Excentre oppposite to B = \(\left(\frac{a x_{1}-b x_{2}+c x_{3}}{a-b+c}, \frac{a y_{1}-b y_{2}+c y_{3}}{a-b+c}\right)\)
  • I3 = Excentre opposite to C = \(\left(\frac{a x_{1}+b x_{2}-c x_{3}}{a+b-c}, \frac{a y_{1}+b y_{2}-c y_{3}}{a+b-c}\right)\)

Ortho Centre Of A Triangle:
Altitude: The line passing through vertex and perpendicular to opposite side of a triangle is called an altitude of the triangle. Altitudes of a triangle are concurrent. The point of concurrence is called the ortho centre of the triangle. It is denoted by “O” or H’.

Circum centre of a Triangle:
Perpendicular bisector: The line passing through mid point of a side and perpendicular to the side is called the perpendicular bisector of the side.
The perpendicular bisectors of the sides of a triangle are concurrent. The point of concurrence is called the circurn centre or the triangle. It is denoted by S.

→ We have mentioned that the Coordinate Geometry unifies the ideas of Algebra and Geometry. Now, in this chapter, we study a set of points, which satisfies certain geometric conditions that can also be represented in the form of algebraic equation.

Definition of Locus :
Let us call a set of geometric conditions ‘consistent’ if there is atleast one point satisfying that set of conditions. For example when A = (1, 0) and B = (3,0), the condition ‘the sum of distances of a point P from A and B is equal to 2’ is consistent, where as the condition ‘the sum of distances of a point Q from A and B is equal to 1’ is not consistent, because there is no point Q such that QA + QB = 1 (since AB = 2)
By the locus of a point we mean the set of all positions that it can take when it is subjected to certain consistent geometric conditions.
Or
Locus:

  • The set of all points (and only those points) which satisfy the given geometrical condition(s) (or properties) is called a locus.
    Eg. The set of points in a plane which are at a constant distance r from a given point C is a locus. Here the locus a circle.
  • The set of points in a plane which are equidistant from two given points A and B is a locus. Here the locus is a straight line and it is the perpendicular bisector of the line segment joining A and B.

Inter 1st Year Maths 1B Locus Formulas

Equation of Locus :
It is clear that, every point on the locus satisfies the given conditions and every point which satisfies the given conditions lies on the locus.
By the equation of a locus we mean an algebraic description of the locus. It is obtained by translating the geometric conditions satisfied by the points on the locus, into equivalent algebraic conditions.

Algebraic descriptions give rise to algebraic equations which sometimes contain more than what is required by the geometric conditions. Thus locus may be a part of the curve represented by the algebraic equation. Usually we call this algebraic equation as the equation of locus. However, to get the full description of the locus, the exact part of the curve, points of which satisfy the given geometric description, must be specified.
or
Equation of a Locus:
An equation f(x, y) = O is said to be the equation of a locus S if every point of S satisfies f(x, y) = O and every point that satisfies f(x, y) = O belongs to S.

An equation of a locus is an algebraic description of the locus. This can be obtained in the following way

  • Consider a point P(x, y) on the locus
  • Write the geometric condition(s) to bc satisfied by P in terms of an equation or in equation in symbols.
  • Apply the proper formula of coordinate geometry and translate the geometric condition(s) into an algebraic equation.
  • Simplify the equation so that it is free from radicals. The equation thus obtained is the required equation of locus.