Area of Triangle
Let (x_{1}, y_{1}), (x_{2}, y_{2}) and (x_{3}, y_{3}) respectively be the coordinates of the vertices A, B, C of a triangle ABC. Then the area of triangle ABC, is
...(1)
....(2)
While using formula (1) or (2). order of the points (x_{l}, y_{L}), (x_{2}, y_{2}) and (x_{3}, y_{3}) has not been taken into account. If we plot the points A(x_{1}, y_{1}), B (x_{2}, y_{2}) and C(x_{3}, y_{3}), then the area of the triangle as obtained by using formula (1) or (2) will be positive or negative as the points A, B, C are in anticlockwise or clockwise directions.
while finding the area of triangle ABC, we use the formula :
Area of ΔABC = Modulus of
Notes
(i) If three points (x_{1}, y_{1}), (x_{2}, y_{2}) and (x_{3}, y_{3}) are collinear, then
(ii) Equation of straight line passing through (x_{1}, y_{1}) and (x_{2}, y_{2}) is given by
(iii) In case of polygon with (x_{1}, y_{1}), (x_{2}, y_{2}) ............ (x_{n,} y_{n}) the area is given by
1/2  (x_{1}y _{2} – y_{1}x _{2} ) + (x_{ 2} y_{3} – y_{2}x_{3} ) + ......... + (x_{n–1}y_{n} – y_{n–1}x_{n} ) + (x_{n} y_{1} – y_{n} x_{1}) 
Ex.3 The vertices of quadrilateral in order are (–1, 4), (5, 6), (2, 9) and (x, x2). The area of the quadrilateral 15/2 sqft. units, then find the point (x, x^{2})
Sol. Area of quadrilateral
3x^{2} – 5x + 7 = ± 15
3x^{2} – 5x – 8 = 0,
3x^{2} – 5x + 22 = 0 ⇒ x = 8/3 , x = –1
Hence point is (8/3,64/9) or (–1, 1). But (–1, 1) will not form a quadrilateral as per given order of the points. Hence the required point is (8/3,64/9)
(1) Locus : When a point moves in a plane under certain geometrical conditions, the point traces out a path. This path of a moving point is called its locus.
Note : All those points which satisfy the given geometrical condition will definitely lie on the locus. But converse is not true always.
(2) Equation of Locus : The equation to a locus is the relation which exists between the coordinates of any point on the path, and which holds for no other point except those lying on the path.
(3) Procedure for finding the equation of the locus of a point
(i) If we are finding the equation of the locus of a point P, assign coordinates (h, k) to P.
(ii) Express the given conditions as equations in terms of the known quantities to facilitate calculations. We sometimes include some unknown quantities known as parameters.
(iii) Eliminate the parameters. so that the eliminate contains only h, k and known quantities.
(iv) Replace h by x, and k by y, in the eliminate. The resulting would be the equation of the locus of P. (v) If x and y coordinates of the moving point are obtained in terms of a third variable t (called the parameter), eliminate t to obtain the relation in x and y and simplify this relation. This will give the required equation of locus.
Ex. Find the focus of the middle points of the segment of a line passing through the point of intersection of the lines ax + by + c = 0 and lx + my + n = 0 and intercepted between the axes.
Sol. Any line (say L = 0) passing through the point of intersection of ax + by + c = 0 and lx + my + n = 0 is (ax + by + c) + λ (lx + my + n) = 0, where l is any real number.
Point of intersection of L = 0 with axes are
Let the mid point be (h, k).
Eliminating λ The required locus is : 2(am – lb) = (lc – an) x + (nb – mc)y..
(4) Inclination of a line : Its a measure of the smallest nonnegative angle which the line makes with +ve direction of the xaxis [angle being measured in anticlockwise direction]. 0 ≤ α < π
(5) Slope of the line : If the inclination of line is θ and then its slope is defined as tan θ and denoted by 'm'
(i) If θ = 0, then m = 0 i.e. line parallel to xaxis.
(ii) If θ = 90º, then m does not exist i.e. line parallel to yaxis
(iii) Slope of line joining two points A(x_{1}, y_{1}) & B(x_{2}, y_{2}) is
(iv) If a line equally inclined with coordinate axes then slope is ± 1.
(6) Intercepts : The point where a line cuts the xa xis (or ya xis ) is called its xintercep t (or yintercept).
(i) Intercepts may be +ve, –ve or zero.
(ii) A line making an intercept of –a withyaxis means the line passing through (0,–a)
(iii) A line makes equal nonzero intercept with both coordinate axes then slope is –1
(iv) A line makes nonzero intercept with both coordinate axes equal in magnitude then slope is ±1.
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1. What is the formula to find the area of a triangle? 
2. How do you find the height of a triangle if it is not given? 
3. Can you find the area of a triangle if only the lengths of the sides are given? 
4. Is it possible to find the area of a triangle if only the angles are given? 
5. How do you find the area of an equilateral triangle? 

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