Learning these TS Inter 1st Year Maths 1B Formulas Chapter 1 Locus will help students to solve mathematical problems quickly.
TS Inter 1st Year Maths 1B Locus Formulas
→ Consider a pair of mutually perpendicular lines of reference X’ X, Y’ Y in a plane. These are called the coordinate axes and their point of intersection is called the origin denoted by ‘O’.
→ Consider a point P in the plane. Let x, denotes the perpendicular distance of P from Y – axis and yx denotes the perpendicular distance of P from X – axis. Then P is represented as ordered pair in the following quadrants.
1st Quadrant → P (x1, y1)
2ndQuadrant → P(-x1, y1)
3rd Quadrant → P (- x1, – y1)
4th Quadrant → P(x1, -y1)
The first element is called the x – coordinate (abscissa) and the second element is called the y- coordinate (ordinate).
- The distance between the points P(x1, y1) and Q(x2, y2) in the plane denoted by
PQ = \(\sqrt{\left(x_1-x_2\right)^2+\left(y_1-y_2\right)^2}=\sqrt{(\text { difference of } x \text {-coordinates })^2+(\text { difference of } y \text {-coordinates })^2}\) - The distance of P(x1, y1) from the origin (0, 0) is OP = \(\sqrt{\mathrm{x}_1^2+\mathrm{y}_1^2}\)
- The distance between [joints A (x1, 0) and B (x2,0) is AB = \(\sqrt{\left(x_1-x_2\right)^2+(0-0)^2}\) = (x1 – x2).
- The distance between points A (0, y1) and B (0, y2) is y1 – y2.
→ Section Formulae:
- The coordinates of the point ‘P’ which divides the line segment joining points A(x1, y1) and B(x2, y2) internally in the ratio m1 : m2 is \(\left(\frac{m_1 x_2+m_2 x_1}{m_1+m_2}, \frac{m_1 y_2+m_2 y_1}{m_1+m_2}\right)\)
- The coordinates of the point P’ which divides the line segment joining points A(x2, y2) and B(x2, y2)Externally in the ratio m1 : m2 is \(\left(\frac{m_1 x_2-m_2 x_1}{m_1-m_2}, \frac{m_1 y_2-m_2 y_1}{m_1-m_2}\right)\)
- Coordinates of midpoint of the line segment \(\overline{\mathrm{AB}}\) is \(\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)\)
- The points which divide a line segment in the ratio 1 : 2 or 2 : 1 are called the points of trisection.
→ (i) The area of the triangle with vertices A(x1, y1), B(x2, y2) and C(x3, y3) is given by
Δ = \(\frac{1}{2}\)|x1(y2 – y3) + x2(y3 – y1)+ x3(y1 – y2) = \(\frac{1}{2}\)Σx1(y2 – y3)
(ii) The area of the triangle OAB with vertices O (0, 0), A(x1, y1) and B(x2, y2) is given by
Δ = \(\frac{1}{2}\)|x1y2 – x2y1|
→ The area of the quadrilateral with vertices A(x1, y1), B(x2, y2) C(x3, y3) and D(x4, y4) is
= \(\frac{1}{2}\)|x1(y2 – y4) + x2(y3 – y1) + x3(y4 – y2) + x4(y1 – y3) = \(\frac{1}{2}\)Σx1(y2 – y4)
→ In a triangle the line segment joining a vertex to the midpoint of opposite side is called the median. Point of intersection of the 3 medians of the triangle is called the centroid of the triangle denoted by G. This point G divides every median internally in the ratio 2 : 1.
→ The coordinates of centroid of the triangle having vertices A (x1, y1), B ( x2, y2) and C (x3, y3) is G = \(\)
→ The bisectors of internal angles of a triangle are concurrent and the point of concurrence is called in center of the triangle denoted by 1.
This is equidistant from three sides and this distance is called the in radius denoted by r. The circle drawn with I as centre and r as radius touches all the three sides internally and this circle is called the in- circle.
→ If A (x1, y1), B ( x2, y2) and C (x3, y3) are the vertices and a,b, and c are respectively the sides BC, CA and AB of triangle ABC, then the coordinates of the in center are
I = \(\left(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}\right)\)
→ In a triangle, one internal angular bisector and two externa! angular bisectors are concurrent j and the point of concurrence is called the Ex-centre of the triangie denoted by I1.
This is equidistant from one side and extensions of the other sides. This distance is called i the Ex-radius of the lriangie denoted by r1. The circle drawn with I1 as centre and r1 as radius touches these sides. This circle is called the Ex-circle of the triangle opposite to the vertex A. Similarly we have two more centres I2 and I3 with radii r2 r3. Coordinates of ex-centres of the triangle are given by
I1 = \(\left(\frac{-a x_1+b x_2+c x_3}{-a+b+c}, \frac{-a y_1+b y_2+c y_3}{-a+b+c}\right)\)
I2 = \(\left(\frac{a x_1-b x_2+c x_3}{a-b+c}, \frac{a y_1-b y_2+c y_3}{a-b+c}\right)\)
I3 = \(\left(\frac{a x_1+b x_2-c x_3}{a+b-c}, \frac{a y_1+b y_2-c y_3}{a+b-c}\right)\)
where A (x1, y1), B ( x2, y2), C (x3, y3) are vertices of the triangle.
→ A set of geometric conditions is said to be consistent if there exists atleast one point satisfying the set of conditions. As an example, if A = (1, 0) and B = (3, 0) then PA + PB = 2 represents the sum of the distances of a point P from A and B is equal to 2, is a consistent condition whereas the distances of a point Q from A and B. ie., QA + QB = 1 is not consistent (∵ AB = 2).
→ Locus is the set of points (and only those points) that satisfy the given consistent geometric conditions. Hence
- Every point satisfying the given condition (s) is a point on the locus.
- Every point on the locus satisfies the given conditions).
→ Equation of locus of a point is an algebraic equation in ‘x’ and y’ satisfied by the points (x, y) on the locus alone. To get the full description of the locus, the exact part of the curve, the points of which satisfy the given geometric description need to be specified.