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CBSE Class 10 Maths Formulas Chapter-07 | Coordinate Geometry
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Introduction
It is that branch of mathematics in which we solve the geometrical problems algebraically.In coordinate geometry we combine geometry and algebra.To locate a point in the plane we use a system of two lines which are perpendicular to each other. This system is called the cartesian coordinate system
Cartesian coordinate
system In this system there are two lines which intersect each other at a point perpendicularly. Point of intersection of two lines is called Origin. Horizontal line is called x-axis and Vertical line is called y-axis. Numbers are written on both the axis on either side of the origin. On x-axis positive numbers are written on the right side of the origin and negative numbers are written on the left side of the origin. On y-axis positive numbers are written above the origin and negative numbers are written below the origin. |
Coordinates of a point
Coordinates of any point in the Cartesian plane can be written as (x, y) or (a, b). First coordinate is the value on the x-axis (or abscissa) and second is the value on the y-axis (or ordinate). Coordinates of the origin is always (0, 0) Coordinates of any point on the x-axis is of the form (x, 0) or (a, 0) Coordinates of any point on the y-axis is of the form (0, y) or (0, b) |
In this blog we will discuss the following point
1) Distance between
two points. 2) Different types of
quadrilateral. 3) Section formula. 4) Mid- point formula. 5) Area of a triangle. 6) Collinearity of
three points. 7) Centroid of a
triangle. 8) Area of
quadrilateral. |
Distance between two points or Distance formula
Let us suppose that two points P(x_{1}, y_{1}) and Q(x_{2}, y_{2}) in the cartesian plane. Then distance between P and Q is denoted by |PQ| and is given by\[|PQ|=\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}\]or\[|PQ|=\sqrt{(Difference \; of \; abscissa)^{2}+(Difference \; of\; ordinate)^{2}}\]
Collinearity of three points by using distance formula
Three points A, B and C are collinear if all these points lie on the straight line.In order to prove that A, B, C are collinear we shall prove that|AB| + |BC| = |AC|
Types of Quadrilaterals By using Distance formula
Let ABCD is a quadrilateral. Then with the help of distance formula find the length of sides |AB|, |BC|, |CD|, |DA| and diagonals |AC| and |BD| i) If opposite sides are equal and diagonals are not equal then quadrilateral is a parallelogram. ii) If opposite sides are equal and diagonals are also equal then quadrilateral is a Rectangle. iii) If all sides are equal and diagonals are not equal then quadrilateral is rhombus. iv) If all sides are equal and diagonals are also equal then quadrilateral is square. v) If two pair of adjacent sides are equal and diagonals are not equal then quadrilateral is a Kite. |
Section Formula [Used in NCERT Ex-7.2]
It is the formula to find the coordinates of the point which divide the given line in the given ratio.Let us suppose that two points A(x_{1}, y_{1}) and B(x_{2}, y_{2}) on a line. Let point P(x, y) divide line segment AB in the ratio m_{1} : m_{2}Therefore by section formula the coordinates of point P(x, y) is given by\[P(x,y)=\left (\frac{m_{1}x_{2}+m_{2}x_{1}}{m_{1}+m_{2}},\; \frac{m_{1}y_{2}+m_{2}y_{1}}{m_{1}+m_{2}} \right )\]Mid-Point FormulaWhen ratio become equal (or m_{1} = m_{2} ) then the point P(x, y) is called the mid-point of the line segment AB and is given by\[P(x,y)=\left (\frac{x_{1}+x_{2}}{2},\frac{y_{1}+y_{2}}{2} \right )\]
Area of Triangle and Quadrilateral [Used in NCERT Ex- 7.3]
Let A(x_{1}, y_{1}), B(x_{2}, y_{2}), C(x_{3}, y_{3}) are the three vertices of a triangle then area of triangle is given as follows\[Area of\Delta =\frac{1}{2}\times | x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1}+x_{2}(y_{1}-y_{2}))|\]Another MethodPlace all the points of the triangle as shown in the figure and then follow the method as shown below\[Area of\Delta =\frac{1}{2}\times |x_{1}y_{2}+x_{2}y_{3}+x_{3}y_{1}-(y_{1}x_{2}+y_{2}x_{3}+y_{3}x_{1}) |\]Area of triangle always be taken positive. If your answer is negative then also convert it into positive.Collinearity of three pointsThree points are collinear if area of triangle equal to zero.Centroid of a triangleThe point of concurrence of all the medians of a triangle is called its centroid.If point P(x, y) is the centroid of triangle ABC then \[P(x,y)=\left ( \frac{x_{1}+x_{2}+x_{3}}{3},\; \frac{y_{1}+y_{2}+y_{3}}{3} \right )\]Median of triangle :A line segment joining the vertex of the triangle with the mid-point of the opposite side is called medianCentroid divide the median in 2:1 and 2 always point towards the vertex of the triangle.Area of quadrilateralPlace all the points of the quadrilateral as shown in the figure above and then follow the method as shown belowArea of quadrilateral \[=\frac{1}{2}\times |x_{1}y_{2}+ x_{2}y_{3}+ x_{3}y_{4}+ x_{4}y_{1}-(y_{1}x_{2}+y_{2}x_{3}+y_{3}x_{4}+y_{4}x_{1}) |\]Area of a quadrilateral is always be taken positive. If your answer is negative then also convert it into positive.
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