Coordinate Geometry — Class 10 Maths Notes

Topics covered in Coordinate Geometry

1. 1.Introduction of Coordinate Geometry

   Coordinate geometry, also known as analytic geometry, is a branch of geometry where the position of points on the plane is described using ordered pairs of numbers (coordinates). The basic idea is to define every point in terms of its distance from two intersecting lines, usually referred to as axes.

   Here’s a simple breakdown of the core concepts in coordinate geometry:

   1. Cartesian Coordinates System:

   • Axes: In a 2-dimensional space, we have two perpendicular lines: the horizontal line is called the x-axis, and the vertical line is called the y-axis.
   • Origin: The point where these axes intersect is called the origin, denoted as (0,0).
   • Coordinates: Any point in this plane can be represented by an ordered pair of numbers (x, y), where 'x' is the distance from the y-axis, and 'y' is the distance from the x-axis.

   2. Quadrants:

   • The plane is divided into four quadrants by the x and y axes.
   • Quadrant I: Both x and y are positive.
   • Quadrant II: x is negative, y is positive.
   • Quadrant III: Both x and y are negative.
   • Quadrant IV: x is positive, y is negative.

   3. Distance Between Two Points:

   • Given two points, P1 (x1, y1) and P2 (x2, y2), the distance 'd' between them can be found using the formula: =(2−1)2+(2−1)2d=(x2−x1)2+(y2−y1)2​

   4. Midpoint of a Line Segment:

   • The midpoint of a segment connecting two points, P1 (x1, y1) and P2 (x2, y2), is given by: =(1+22,1+22)M=(2x1+x2​,2y1+y2​)

   5. Slope of a Line:

   • The slope (m) of a line connecting two points, P1 (x1, y1) and P2 (x2, y2), indicates its steepness and direction: =2−12−1m=x2−x1y2−y1​

   Real-Life Examples: 1. Imagine you’re using a ride-sharing app to go from your home to a restaurant. The app calculates the best route using coordinate geometry by taking your current location and the restaurant's coordinates to guide you there.

   2. Imagine you're in a city where the streets are laid out in a grid pattern. Your home is located at the point (2,3) which means it's 2 blocks east and 3 blocks north of the city center (the origin). If your school is at (5,7), then the coordinate geometry can help you find the shortest path to school and the exact distance you'd travel.

   Numerical Example:

   Let’s find the distance between your home (2,3) and school (5,7). Using the distance formula: =(5−2)2+(7−3)2d=(5−2)2+(7−3)2​ =32+42d=32+42​ =9+16d=9+16​ =25d=25​ =5d=5 So, the school is 5 units of distance from your home on this grid.

   Real-Life Applications:

   1. Navigation and Mapping: GPS technology relies on coordinate geometry to pinpoint locations. We use it daily to find destinations using map applications on smartphones.

   2. Architecture: Architects design buildings by creating blueprints with coordinates to specify different locations and dimensions.

   3. Computer Graphics: Whether it’s for creating movies, video games, or simulations, coordinate geometry is used to render shapes and movements on screens.

   4. Astronomy: Astronomers locate stars and planets by their coordinates in the sky, which helps in exploration and study of the universe.

   5. Robotics: Robots are programmed to understand their position and navigate using coordinate systems.

   Career Fields:

   1. Surveying: Surveyors use coordinate geometry to measure land and create maps.

   2. Civil Engineering: Civil engineers use it for infrastructure development, including roads, bridges, and dams.

   3. Aerospace: This field uses coordinate geometry to plot flight paths for aircraft and spacecraft.

   4. Urban Planning: Planners use it to design layouts for developments, transportation systems, and public spaces.

2. 2.Do the points (3, 2), (–2, –3) and (2, 3) form a triangle? If so, name the type of triangle formed.

   To determine if the points (3, 2), (–2, –3), and (2, 3) form a triangle, we can calculate the distances between each pair of points. If all three distances are non-zero and the sum of any two distances is greater than the third distance, the points form a triangle.

   Let's label the points A(3, 2), B(–2, –3), and C(2, 3) and calculate the distances AB, BC, and AC using the distance formula:

   Distance between two points (1,1)(x1​,y1​) and (2,2)(x2​,y2​) is given by: Distance=(2−1)2+(2−1)2Distance=(x2​−x1​)2+(y2​−y1​)2​

   Distance AB: =(−2−3)2+(−3−2)2AB=(−2−3)2+(−3−2)2​ =(−5)2+(−5)2AB=(−5)2+(−5)2​ =25+25AB=25+25​ =50AB=50​ =52AB=52​

   Distance BC: =(2−(−2))2+(3−(−3))2BC=(2−(−2))2+(3−(−3))2​ =42+62BC=42+62​ =16+36BC=16+36​ =52BC=52​ =213BC=213​

   Distance AC: =(2−3)2+(3−2)2AC=(2−3)2+(3−2)2​ =(−1)2+(1)2AC=(−1)2+(1)2​ =1+1AC=1+1​ =2AC=2​ =2AC=2​

   Since all three distances are non-zero, we can confirm that the points do indeed form a triangle.

   To determine the type of triangle, we need to check if it is a right-angled triangle by using the Pythagorean theorem:

   2+2=?2AB2+AC2=?BC2 (52)2+(2)2=?(213)2(52​)2+(2​)2=?(213​)2 50+2=?5250+2=?52

   Since 50+2=5250+2=52, the Pythagorean theorem holds true for the distances between points A, B, and C. Hence, the points form a right-angled triangle with AB as the hypotenuse.

3. 3.Find a relation between x and y such that the point (x , y) is equidistant from the points (7, 1) and (3, 5).

   To find the relation between x and y such that the point (,)(x,y) is equidistant from the points (7,1)(7,1) and (3,5)(3,5), we need to set the distance from (,)(x,y) to both points equal.

   The distance between two points (1,1)(x1​,y1​) and (2,2)(x2​,y2​) is given by the formula:

   Distance=(2−1)2+(2−1)2Distance=(x2​−x1​)2+(y2​−y1​)2​

   Let's calculate the distance of (,)(x,y) from both (7,1)(7,1) and (3,5)(3,5) and set them equal.

   For the point (7,1)(7,1):

   1=(−7)2+(−1)2d1​=(x−7)2+(y−1)2​

   For the point (3,5)(3,5):

   2=(−3)2+(−5)2d2​=(x−3)2+(y−5)2​

   Since 1d1​ is equidistant to 2d2​, we have:

   (−7)2+(−1)2=(−3)2+(−5)2(x−7)2+(y−1)2​=(x−3)2+(y−5)2​

   Squaring both sides to remove the square roots, we get:

   (−7)2+(−1)2=(−3)2+(−5)2(x−7)2+(y−1)2=(x−3)2+(y−5)2

   Expanding and simplifying both sides:

   2−14+49+2−2+1=2−6+9+2−10+25x2−14x+49+y2−2y+1=x2−6x+9+y2−10y+25

   −14+50−2=−6+34−10−14x+50−2y=−6x+34−10y

   Bringing like terms to one side:

   −14+6+2−10=34−50−14x+6x+2y−10y=34−50

   −8+8=−16−8x+8y=−16

   Dividing by -8:

   −=2x−y=2

   So the relation between x and y such that the point (,)(x,y) is equidistant from (7,1)(7,1) and (3,5)(3,5) is −=2x−y=2.

4. 4.Find a point on the y-axis which is equidistant from the points A(6, 5) and B(– 4, 3).

   To find a point on the y-axis that is equidistant from points A(6, 5) and B(–4, 3), we need to remember that any point on the y-axis has an x-coordinate of 0. So, the point we are looking for will be of the form (0, y).

   The distance from a point (x, y) to A(6, 5) and B(–4, 3) can be given by the distance formula:

   Distance=(2−1)2+(2−1)2Distance=(x2​−x1​)2+(y2​−y1​)2​

   Let's call the point on the y-axis P(0, y).

   Distance from P to A:

   =(6−0)2+(5−)2dPA​=(6−0)2+(5−y)2​ =36+(5−)2dPA​=36+(5−y)2​

   Distance from P to B:

   =(−4−0)2+(3−)2dPB​=(−4−0)2+(3−y)2​ =16+(3−)2dPB​=16+(3−y)2​

   Since P is equidistant from A and B, we set =dPA​=dPB​:

   36+(5−)2=16+(3−)236+(5−y)2​=16+(3−y)2​

   Squaring both sides to remove the square root:

   36+(5−)2=16+(3−)236+(5−y)2=16+(3−y)2

   Expand the squared terms:

   36+25−10+2=16+9−6+236+25−10y+y2=16+9−6y+y2

   Now, we can cancel 2y2 from both sides and simplify:

   36+25−10=16+9−636+25−10y=16+9−6y 61−10=25−661−10y=25−6y

   Bringing like terms to one side gives us:

   4=61−254y=61−25 4=364y=36

   Dividing by 4:

   =9y=9

   Therefore, the point on the y-axis that is equidistant from A(6, 5) and B(–4, 3) is (0, 9).

5. 5.Find the distance between the following pairs of points : (i) (2, 3), (4, 1), (ii) (– 5, 7), (– 1, 3), (iii) (a, b), (– a, – b)

   To find the distance between two points in a plane, we use the distance formula derived from the Pythagorean theorem. The distance (d) between two points (1,1)(x1​,y1​) and (2,2)(x2​,y2​) is given by:

   =(2−1)2+(2−1)2d=(x2​−x1​)2+(y2​−y1​)2​

   Let's calculate the distance for each pair of points:

   (i) For the points (2,3)(2,3) and (4,1)(4,1):

   1=(4−2)2+(1−3)2d1​=(4−2)2+(1−3)2​ 1=22+(−2)2d1​=22+(−2)2​ 1=4+4d1​=4+4​ 1=8d1​=8​ 1=22d1​=22​

   (ii) For the points (−5,7)(−5,7) and (−1,3)(−1,3):

   2=(−1+5)2+(3−7)2d2​=(−1+5)2+(3−7)2​ 2=42+(−4)2d2​=42+(−4)2​ 2=16+16d2​=16+16​ 2=32d2​=32​ 2=42d2​=42​

   (iii) For the points (,)(a,b) and (−,−)(−a,−b):

   3=(−−)2+(−−)2d3​=(−a−a)2+(−b−b)2​ 3=(−2)2+(−2)2
   ,d3​=(−2a)2+(−2b)2,​ 3=42+42d3​=4a2+4b2​ 3=4(2+2),d3​=4(a2+b2)​ 3=22+2,d3​=2a2+b2​
   
   

6. 6.Distance Formula

   The Distance Formula is used to find out how far two points are from each other on a graph. Imagine you have two dots on a piece of graph paper and you want to know the straight line distance between them. This formula helps you calculate that exact distance without having to measure it with a ruler.

   In simple terms, if you have two points, let's call them Point A and Point B, and they have coordinates (x₁, y₁) and (x₂, y₂) respectively, the Distance Formula looks like this: Distance=(2−1)2+(2−1)2 Distance=(x2​−x1​)2+(y2​−y1​)2​ This formula is just a version of the Pythagorean theorem that you might have learned about in triangles.

   Numerical Example: Let's say Point A is at (3, 4) and Point B is at (7, 1). The distance between them would be calculated as: Distance=(7−3)2+(1−4)2Distance=(7−3)2+(1−4)2​ Distance=(4)2+(−3)2Distance=(4)2+(−3)2​ Distance=16+9Distance=16+9​ Distance=25Distance=25​ Distance=5Distance=5 So, the distance between Point A and Point B is 5 units.

   Real-Life Examples and Where it is apply in Careers:

   • In navigation, like when sailors or pilots calculate the shortest path between two points.
   • Architects use it to find distances in their plans without needing to measure physically each time.
   • Game developers use it for programming how characters move or how objects interact based on their distances.
   • Surveyors rely on this formula to determine property boundaries.

7. 7.Determine if the points (1, 5), (2, 3) and (– 2, – 11) are collinear.

   To determine if three points are collinear, we check if the slopes between each pair of points are equal. If the slopes are equal, then the points lie on the same line and are therefore collinear.

   The slope between two points (1,1)(x1​,y1​) and (2,2)(x2​,y2​) is given by the formula:

   Slope=2−12−1Slope=x2​−x1​y2​−y1​​

   Let's calculate the slopes between the pairs of points:

   For points (1, 5) and (2, 3):

   Slope1−2=3−52−1=−21=−2Slope1−2​=2−13−5​=1−2​=−2

   For points (2, 3) and (–2, –11):

   Slope2−3=−11−3−2−2=−14−4=3.5Slope2−3​=−2−2−11−3​=−4−14​=3.5

   Since the slopes Slope1−2Slope1−2​ and Slope2−3Slope2−3​ are not equal, the points (1, 5), (2, 3), and (–2, –11) are not collinear.

8. 8.Check whether (5, – 2), (6, 4) and (7, – 2) are the vertices of an isosceles triangle.

   To determine whether three points form an isosceles triangle, we need to check if any two sides of the triangle are of equal length.

   Let's name the points (5,−2)A(5,−2), (6,4)B(6,4), and (7,−2)C(7,−2).

   We'll calculate the distance between each pair of points using the distance formula:

   =(2−1)2+(2−1)2d=(x2​−x1​)2+(y2​−y1​)2​

   For AB:

   =(6−5)2+(4−(−2))2=12+62=1+36=37dAB​=(6−5)2+(4−(−2))2​=12+62​=1+36​=37​

   For BC:

   =(7−6)2+(−2−4)2=12+(−6)2=1+36=37dBC​=(7−6)2+(−2−4)2​=12+(−6)2​=1+36​=37​

   For AC:

   =(7−5)2+(−2−(−2))2=22+02=4+0=4=2dAC​=(7−5)2+(−2−(−2))2​=22+02​=4+0​=4​=2

   Since =AB=BC and both are different from AC, the triangle formed by these points is indeed an isosceles triangle with AB and BC as the equal sides.

9. 9.Find a point on the y-axis which is equidistant from the points A (6, 5) and B(– 4, 3).

   To find a point on the y-axis that is equidistant from points A(6, 5) and B(–4, 3), we are essentially looking for a point with coordinates (0, y) because any point on the y-axis will have an x-coordinate of 0.

   The distance from (0, y) to A(6, 5) and from (0, y) to B(–4, 3) should be equal. We can use the distance formula for this: =(2−1)2+(2−1)2d=(x2​−x1​)2+(y2​−y1​)2​.

   For point A(6, 5):

   =(6−0)2+(5−)2dA​=(6−0)2+(5−y)2​ =36+(5−)2dA​=36+(5−y)2​

   For point B(–4, 3):

   =(−4−0)2+(3−)2dB​=(−4−0)2+(3−y)2​ =16+(3−)2dB​=16+(3−y)2​

   Since =dA​=dB​, we can set the expressions equal to each other and solve for y:

   36+(5−)2=16+(3−)236+(5−y)2​=16+(3−y)2​

   Squaring both sides to remove the square root gives us:

   36+(5−)2=16+(3−)236+(5−y)2=16+(3−y)2

   Expanding the squared terms:

   36+25−10+2=16+9−6+236+25−10y+y2=16+9−6y+y2

   Now, we simplify and solve for y:

   61−10=25−661−10y=25−6y 36=436=4y =9y=9

   So, the point on the y-axis equidistant from points A and B is (0, 9).

10. 10.Find the point on the x-axis which is equidistant from (2, –5) and (–2, 9).

    To find the point on the x-axis that is equidistant from (2, -5) and (-2, 9), we need to remember that any point on the x-axis has a y-coordinate of 0. Let's call the point we're looking for P(x, 0).

    The distance between two points (x1, y1) and (x2, y2) in a plane is given by the distance formula:

    Distance=(2−1)2+(2−1)2Distance=(x2−x1)2+(y2−y1)2​

    The point P(x, 0) must be equidistant from (2, -5) and (-2, 9), so the distances between P and each of these points must be equal.

    Let's set up the equation using the distance formula:

    For the point (2, -5): (−2)2+(0−(−5))2(x−2)2+(0−(−5))2​

    For the point (-2, 9): (−(−2))2+(0−9)2(x−(−2))2+(0−9)2​

    Now, we set these two expressions equal to each other because the distances are the same:

    (−2)2+52=(+2)2+92(x−2)2+52​=(x+2)2+92​

    Squaring both sides to remove the square roots gives us:

    (−2)2+25=(+2)2+81(x−2)2+25=(x+2)2+81

    Let's solve this equation for x:

    2−4+4+25=2+4+4+81x2−4x+4+25=x2+4x+4+81

    Combine like terms:

    2−4+29=2+4+85x2−4x+29=x2+4x+85

    Subtract 2x2 from both sides:

    −4+29=4+85−4x+29=4x+85

    Add 4x to both sides:

    29=8+8529=8x+85

    Subtract 85 from both sides:

    −56=8−56=8x

    Divide by 8:

    =−7x=−7

    So, the point on the x-axis equidistant from (2, -5) and (-2, 9) is (-7, 0).

11. 11.If Q(0, 1) is equidistant from P(5, –3) and R(x, 6), find the values of x. Also find the distances QR and PR.

    To find the value of x such that the point (0,1)Q(0,1) is equidistant from (5,−3)P(5,−3) and (,6)R(x,6), we will use the distance formula as before.

    The distance between two points (1,1)(x1​,y1​) and (2,2)(x2​,y2​) is given by:

    Distance=(2−1)2+(2−1)2Distance=(x2​−x1​)2+(y2​−y1​)2​

    Since Q is equidistant from P and R, we can set the distances QP and QR equal to each other.

    Distance QP is:

    =(5−0)2+(−3−1)2QP=(5−0)2+(−3−1)2​ =52+(−4)2QP=52+(−4)2​ =25+16QP=25+16​ =41QP=41​

    Distance QR is:

    =(−0)2+(6−1)2QR=(x−0)2+(6−1)2​ =2+52QR=x2+52​ =2+25QR=x2+25​

    Now, we set =QP=QR:

    41=2+2541​=x2+25​

    Square both sides to eliminate the square root:

    41=2+2541=x2+25

    Subtract 25 from both sides:

    16=216=x2

    Take the square root of both sides:

    =±4x=±4

    So, there are two points R on the line =6y=6 that are equidistant from Q: (4,6)R(4,6) and (−4,6)R(−4,6).

    Now, we already calculated QP which is 4141​. The distances QR and PR will be the same because Q is equidistant from P and R.

12. 12.Find a relation between x and y such that the point (x, y) is equidistant from the point (3, 6) and (– 3, 4).

    To find a relation between x and y such that the point (,)(x,y) is equidistant from the points (3,6)(3,6) and (−3,4)(−3,4), we will use the distance formula for each pair of points and then set the two expressions equal to each other.

    The distance formula between two points (1,1)(x1​,y1​) and (2,2)(x2​,y2​) is given by:

    Distance=(2−1)2+(2−1)2Distance=(x2​−x1​)2+(y2​−y1​)2​

    Let 1D1​ be the distance between (,)(x,y) and (3,6)(3,6), and 2D2​ be the distance between (,)(x,y) and (−3,4)(−3,4).

    1=(−3)2+(−6)2D1​=(x−3)2+(y−6)2​ 2=(+3)2+(−4)2D2​=(x+3)2+(y−4)2​

    Since the distances are equal:

    1=2D1​=D2​

    (−3)2+(−6)2=(+3)2+(−4)2(x−3)2+(y−6)2​=(x+3)2+(y−4)2​

    Square both sides to eliminate the square root:

    (−3)2+(−6)2=(+3)2+(−4)2(x−3)2+(y−6)2=(x+3)2+(y−4)2

    Now, expand both sides:

    2−6+9+2−12+36=2+6+9+2−8+16x2−6x+9+y2−12y+36=x2+6x+9+y2−8y+16

    Combine like terms and simplify:

    −6+2−12+45=6+2−8+25−6x+y2−12y+45=6x+y2−8y+25

    Move all terms involving x to one side and y to the other side:

    −6−6=−8+12+25−45−6x−6x=−8y+12y+25−45

    −12=4−20−12x=4y−20

    Divide both sides by -12 to solve for x:

    =−13+53x=−31​y+35​

    So the relation between x and y that ensures (,)(x,y) is equidistant from (3,6)(3,6) and (−3,4)(−3,4) is:

    =−13+53x=−31​y+35​
    
    

13. 13.Quick Revision

    1. Introduction to Coordinate Geometry: Coordinate geometry is a part of mathematics where you use a grid to describe the exact position of points, lines, and shapes. This grid is made up of two lines: the horizontal line (x-axis) and the vertical line (y-axis). Each point on the grid is identified by an ordered pair of numbers, called coordinates, like (x, y).

    2. Distance Formula: The distance formula helps you find the distance between two points on the coordinate grid. If you have two points, point A with coordinates (x₁, y₁) and point B with coordinates (x₂, y₂), the distance (d) between these points is given by the formula: =(2−1)2+(2−1)2d=(x2​−x1​)2+(y2​−y1​)2​

    Implementing the Distance Formula: To use this formula:

    • Take the x-coordinate of the first point and subtract it from the x-coordinate of the second point, then square the result.
    • Do the same with the y-coordinates.
    • Add these two squared numbers together.
    • Take the square root of that sum, and you've found the distance between the two points.