Overview
This chapter introduces Coordinate Geometry (Cartesian system) for Class 10 students and develops analytic methods to study geometric figures on a plane using algebra. It begins with the Cartesian plane, axes, quadrants and plotting points (ordered pairs), then builds toward key formulas — distance between two points, midpoint, section formula (internal and external) and the area of a triangle using coordinates. The chapter emphasizes derivations (so students understand where formulas come from), skillful application to solve numerical and proof-type problems, and connections to graphing linear equations. Coordinate Geometry is important because it links algebra and geometry, provides precise tools for measuring length and area, aids problem solving (collinearity, loci, dividing line segments), and forms the foundation for later topics such as straight lines, circles and conic sections. By the end of the chapter a student will be able to plot points and figures, derive and apply the distance, midpoint and section formulas, compute the area of triangles using determinants, check collinearity, and solve coordinate-based problems accurately and efficiently.
Learning Objectives
- Define the Cartesian coordinate system and explain the role of axes, origin and quadrants.
- Plot points on the Cartesian plane and read off their coordinates from given figures.
- Apply the distance formula to calculate the distance between two points and solve related numerical problems.
- Use the section formula (internal and external) to find coordinates of a point dividing a line segment in a given ratio.
- Determine the coordinates of the midpoint of a segment and apply midpoint results to problem-solving.
- Derive and use the slope formula to find the slope of a line through two given points.
- Identify and write the equation of a straight line in slope-intercept, point-slope, and two-point forms.
- Calculate and use conditions for parallelism and perpendicularity of two lines using their slopes.
Topics in this chapter
7 topics · tap a topic title to jump straight to it.
Cartesian Coordinate System
Definition: The Cartesian coordinate system (or rectangular coordinate system) is a plane divided by two perpendicular number lines: the horizontal axis (x-axis) and the vertical axis (y-axis). Their intersection is the origin O(0,0). Every point in the plane is identified by an ordered pair (x, y) called its coordinates: x is the abscissa (horizontal distance from the origin) and y is the ordinate (vertical distance from the origin).
Main components and conventions:
- Origin: O(0,0).
- Axes: x-axis (y = 0) and y-axis (x = 0).
- Quadrants: Four regions numbered I, II, III, IV counterclockwise from the upper-right. Sign of coordinates in each quadrant: I (+,+), II (−,+), III (−,−), IV (+,−).
- Ordered pair convention: (x, y) — first horizontal then vertical movement. To plot (x, y) move x units along x-axis (right if x>0, left if x<0), then y units parallel to y-axis (up if y>0, down if y<0).
Why it is useful: It converts geometric positions and shapes into algebraic expressions. Points, lines, distances and slopes can be studied using algebraic formulas, enabling precise calculations and graphs.
Basic plotting steps:
- Draw two perpendicular number lines crossing at origin and mark a scale on each axis.
- To plot (x, y): start at origin, move horizontally to x, then vertically to y; mark the point and label it.
- To read coordinates of a point, drop perpendiculars to axes: the intersection values give x and y.
Connections to other topics: Distance between points, midpoint, slope of a line and equations of lines all use the Cartesian system. Many geometric constructions (triangles, circles, polygons) can be analyzed by assigning coordinates to vertices.
- City grid: Streets running east–west and north–south form a Cartesian-like grid. If an intersection is 3 blocks east and 2 blocks north of the city center, its coordinate is (3, 2).
- Computer graphics: Pixel positions on a screen are addressed by (x, y) coordinates (note: some systems use origin at top-left and a downward positive y).
- Robot navigation: A robot on a factory floor can use Cartesian coordinates to move to precise positions (x, y) relative to a fixed origin.
- Plotting points and shapes: Plot A(2,3), B(-3,4), C(-2,-3), D(4,-2) on graph paper to see how points lie in different quadrants and to form shapes by joining them.
- Distance between two points A(x1, y1) and B(x2, y2): d = sqrt((x2 - x1)^2 + (y2 - y1)^2)
- Midpoint of AB (x1, y1) and (x2, y2): M = ((x1 + x2)/2, (y1 + y2)/2)
- Slope (gradient) of line through (x1, y1) and (x2, y2): m = (y2 - y1)/(x2 - x1) (provided x2 != x1)
- Equations of axes and simple lines: x-axis: y = 0; y-axis: x = 0; vertical line through a: x = a; horizontal line through b: y = b
- Distance of point (x, y) from x-axis = |y|; distance from y-axis = |x|
Distance Between Two Points
What it is: In the Cartesian plane, the distance between two points A(x1, y1) and B(x2, y2) is the length of the straight line segment joining them. We derive the formula using the Pythagorean theorem.
Derivation: Draw the horizontal and vertical projections from A and B to form a right-angled triangle. The horizontal leg length = |x2 - x1| and the vertical leg length = |y2 - y1|. By Pythagoras, the distance AB = √[(horizontal leg)2 + (vertical leg)2].
Standard formula:
AB = sqrt((x2 - x1)^2 + (y2 - y1)^2)
Notes and special cases:
- Because squares remove sign, you can write AB = sqrt((x1 - x2)^2 + (y1 - y2)^2) as well.
- If x1 = x2 (same vertical line), AB = |y2 - y1| (vertical distance).
- If y1 = y2 (same horizontal line), AB = |x2 - x1| (horizontal distance).
- Distance of a point P(x, y) from the origin O(0, 0): OP = sqrt(x^2 + y^2).
- Often it is useful to work with squared distance (no square root) when comparing distances: AB^2 = (x2 - x1)^2 + (y2 - y1)^2.
Relation to real problems: This formula is used whenever you need straight-line distance on a coordinate grid—maps, computer graphics, robotics, and measurements in design. (Note: For geographic latitude/longitude on the globe, this flat-plane formula is only an approximation for small regions.)
- Example 1 — Simple integer points: A(1, 2), B(4, 6). Compute AB = sqrt((4-1)^2 + (6-2)^2) = sqrt(3^2 + 4^2) = sqrt(9 + 16) = sqrt(25) = 5.
- Example 2 — Points with negatives: A(-2, 3), B(3, -1). AB = sqrt((3 - (-2))^2 + (-1 - 3)^2) = sqrt(5^2 + (-4)^2) = sqrt(25 + 16) = sqrt(41) ≈ 6.403.
- Example 3 — Same x (vertical distance): A(2, 1), B(2, 6). AB = |6 - 1| = 5 (since x coordinates equal, points lie on a vertical line).
- Example 4 — Distance from origin: P(-3, 4). OP = sqrt((-3)^2 + 4^2) = sqrt(9 + 16) = sqrt(25) = 5.
- Distance between A(x1, y1) and B(x2, y2): AB = sqrt((x2 - x1)^2 + (y2 - y1)^2).
- Horizontal case (y1 = y2): AB = |x2 - x1|.
- Vertical case (x1 = x2): AB = |y2 - y1|.
- Distance from origin O(0,0) to P(x,y): OP = sqrt(x^2 + y^2).
- Squared distance (useful for comparison): AB^2 = (x2 - x1)^2 + (y2 - y1)^2.
- 3D generalisation (for reference): distance between (x1,y1,z1) and (x2,y2,z2) = sqrt((x2-x1)^2 + (y2-y1)^2 + (z2-z1)^2).
Section Formula (Division of a Line Segment)
What it is: If A(x1, y1) and B(x2, y2) are two points in the plane and P is a point on the line AB that divides the segment in a given ratio, the section formula gives the coordinates of P in terms of x1, y1, x2, y2 and the ratio.
Internal division (AP:PB = m:n): The point P that lies on the segment AB and divides it in the ratio m:n (measured from A to B) has coordinates
P = ( (n*x1 + m*x2)/(m + n), (n*y1 + m*y2)/(m + n) ).
Derivation (brief): If AP:PB = m:n then AP = (m/(m+n)) AB. So P = A + (m/(m+n))(B - A). Expanding components produces the formula above.
External division (P lies on the extension of AB, AP:PB = m:n): For a point P that divides AB externally in the ratio m:n,
P = ( (m*x2 - n*x1)/(m - n), (m*y2 - n*y1)/(m - n) ), provided m ≠ n.
Special cases and notes:
- Midpoint: for m = n, internal division gives P = ( (x1 + x2)/2, (y1 + y2)/2 ).
- For center of mass of two point-masses m1 at A and m2 at B, the position is (m1*x1 + m2*x2)/(m1 + m2), (m1*y1 + m2*y2)/(m1 + m2). This is a section formula where the point divides AB internally in the ratio m2:m1 (inverse of masses).
- Keep the order consistent: AP:PB = m:n means the weight m multiplies coordinates of B in the numerator and n multiplies coordinates of A.
Vector/parameter view (useful conceptually): Let t = m/(m + n). Then P = (1 - t)A + tB = ( (1 - t)x1 + t x2, (1 - t)y1 + t y2 ). Varying t from 0 to 1 moves P from A to B.
- Internal division: A(2, 3), B(8, 7), AP:PB = 3:2. Using P = ((n*x1 + m*x2)/(m+n), (n*y1 + m*y2)/(m+n)) with m=3, n=2 gives P = ((2*2 + 3*8)/5, (2*3 + 3*7)/5) = (28/5, 27/5) = (5.6, 5.4).
- External division: A(1, 2), B(5, 4), AP:PB = 2:1 (external). Using P = ((m*x2 - n*x1)/(m - n), (m*y2 - n*y1)/(m - n)) with m=2, n=1 gives P = ((2*5 - 1*1)/(1), (2*4 - 1*2)/(1)) = (9, 6).
- Midpoint: A(1,1), B(5,3). Midpoint = ((1+5)/2, (1+3)/2) = (3, 2).
- Center of mass (real-life): Two masses 2 kg at A(0,0) and 3 kg at B(4,0). Centre = ((2*0 + 3*4)/5, 0) = (12/5, 0) = (2.4, 0). This point divides segment AB internally in the ratio 3:2.
- Internal division (AP:PB = m:n): P = ( (n*x1 + m*x2)/(m + n), (n*y1 + m*y2)/(m + n) )
- External division (AP:PB = m:n): P = ( (m*x2 - n*x1)/(m - n), (m*y2 - n*y1)/(m - n) ), (m ≠ n)
- Midpoint (m = n): P = ( (x1 + x2)/2, (y1 + y2)/2 )
- Parameter form: t = m/(m + n), P = ( (1 - t)*x1 + t*x2, (1 - t)*y1 + t*y2 )
- Center of mass for masses m1 at A and m2 at B: ( (m1*x1 + m2*x2)/(m1 + m2), (m1*y1 + m2*y2)/(m1 + m2) )
Midpoint Formula
Definition: The midpoint of a line segment joining two points A(x₁, y₁) and B(x₂, y₂) in the coordinate plane is the point M that lies exactly halfway between A and B. M divides the segment AB into two equal parts.
Formula and Derivation:
If A(x₁, y₁) and B(x₂, y₂), the midpoint M = (x_m, y_m) is obtained by averaging the corresponding coordinates:
x_m = (x₁ + x₂) / 2, y_m = (y₁ + y₂) / 2
Derivation (idea): the x-coordinate of the midpoint is the average of x₁ and x₂ because the midpoint is equidistant from both x-values; similarly for y. In vector terms M = (A + B)/2.
Properties:
- M lies on segment AB and is equidistant from A and B.
- For a rectangle or any parallelogram, diagonals bisect each other, so their intersection is the midpoint of each diagonal.
- Midpoint is a special case of the section formula (internal division) with ratio 1:1.
Connections: You can use midpoint together with the distance formula to check equal lengths, and with slope to find the perpendicular bisector: its equation passes through M and has slope = −1/(slope of AB) (if AB is not vertical or horizontal).
Special cases:
- Horizontal segment: y₁ = y₂ → midpoint y-coordinate = y₁.
- Vertical segment: x₁ = x₂ → midpoint x-coordinate = x₁.
Real-life uses: navigation (finding halfway point between two places), computer graphics (finding center of a segment), engineering and construction (bisecting beams or distances), GIS (midpoint of two coordinates), robotics (meeting point planning).
- Example 1 (positive coordinates): A(2, 3) and B(8, 7). Midpoint M = ((2+8)/2, (3+7)/2) = (5, 5).
- Example 2 (with negatives): A(-4, 6) and B(2, -2). Midpoint M = ((-4+2)/2, (6+(-2))/2) = (-1, 2).
- Example 3 (vertical segment): A(5, 1) and B(5, 9). Midpoint M = ((5+5)/2, (1+9)/2) = (5, 5).
- Application example: In a rectangle with opposite corners at A(1,2) and C(7,8), the center (intersection of diagonals) is midpoint M = ((1+7)/2, (2+8)/2) = (4,5).
- Midpoint of AB: M = ((x1 + x2)/2, (y1 + y2)/2)
- 1D midpoint on x-axis between x1 and x2: x_m = (x1 + x2)/2
- Section formula (internal division) in ratio m:n gives point P = ((n*x1 + m*x2)/(m+n), (n*y1 + m*y2)/(m+n)); for midpoint m = n = 1 this reduces to the midpoint formula
- Vector form: M = (A + B)/2 = ( (x1 + x2)/2 , (y1 + y2)/2 )
- Perpendicular bisector of AB: passes through M and has slope = -1/m_AB (if slope of AB, m_AB, exists)
Area of a Triangle Using Coordinates
What it is: The area of a triangle with vertices given by coordinates can be computed directly from those coordinates without measuring base and height. This uses the determinant (or "shoelace") method from coordinate geometry.
Main idea / derivation (brief): Take triangle with vertices A(x1,y1), B(x2,y2), C(x3,y3). The area of the parallelogram formed by vectors AB and AC equals the absolute value of their 2D cross product: |(x2-x1)(y3-y1) - (x3-x1)(y2-y1)|. The triangle is half that parallelogram, so:
Area = (1/2) * |(x2 - x1)(y3 - y1) - (x3 - x1)(y2 - y1)|
This is algebraically equivalent to the 3×3 determinant (shoelace) form:
Area = (1/2) * | x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2) |
Notes:
- Use the absolute value because the determinant can be positive or negative depending on the vertex ordering; area is always non‑negative.
- If the computed area = 0, the three points are collinear (lie on a straight line).
- Either of the equivalent formulae may be used—choose the one easiest for calculation.
When to use: Useful for coordinate-plane problems (Class 10), surveying, mapping, computer graphics, and any situation where vertex coordinates are known.
- Example 1 — Direct use of the shoelace formula: A(1,2), B(4,6), C(5,2). Compute area = (1/2)|1(6-2) + 4(2-2) + 5(2-6)| = (1/2)|4 + 0 - 20| = (1/2)*16 = 8 square units.
- Example 2 — Using vector difference form (base along x-axis): A(0,0), B(4,0), C(1,3). Area = (1/2)| (4-0)(3-0) - (1-0)(0-0) | = (1/2)|12 - 0| = 6 square units.
- Example 3 — Collinear check: A(0,0), B(2,2), C(4,4). Area = (1/2)|0(2-4) + 2(4-0) + 4(0-2)| = (1/2)|0 + 8 - 8| = 0 → points are collinear.
- Area = (1/2) * | x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2) | (shoelace / 3×3 determinant form)
- Area = (1/2) * | (x2 - x1)(y3 - y1) - (x3 - x1)(y2 - y1) | (vector / cross-product form)
- Area = (1/2) * | det( [ [x1, y1, 1], [x2, y2, 1], [x3, y3, 1] ] ) |
- Collinear condition: points are collinear ⇔ x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2) = 0
Collinearity of Three Points
Definition: Three points A(x1, y1), B(x2, y2) and C(x3, y3) are said to be collinear if they lie on one straight line.
Two main methods to test collinearity:
- Slope method: If the slopes of AB and AC are equal, the points are collinear. That is, if x2 — x1 and x3 — x1 are non-zero, check
(y2 — y1)/(x2 — x1) = (y3 — y1)/(x3 — x1).
Handle special cases: if AB is vertical (x2 = x1) then AC must also be vertical (x3 = x1) for collinearity. - Determinant (area) method: The area of triangle ABC is zero when the three points are collinear. Use the determinant:
|x1 y1 1|
Expanding gives the condition
|x2 y2 1| = 0
|x3 y3 1|x1(y2 — y3) + x2(y3 — y1) + x3(y1 — y2) = 0.
Equivalently, area = (1/2) * absolute value of that determinant, so area = 0 if and only if points are collinear.
Remarks and special cases:
- If two points coincide (for example A = B) and the third point C is different, all three are considered collinear because the unique line through A(=B) and C contains all given points.
- Horizontal lines: check y-coordinates equal. Vertical lines: check x-coordinates equal. These are handled naturally by the slope or determinant method.
- Alternatively, find equation of line through two points and verify the third satisfies it: line through A(x1,y1) and B(x2,y2) can be written as (y — y1)(x2 — x1) = (x — x1)(y2 — y1); substitute (x3,y3).
How to use in practice (step-by-step):
- Choose a method (slope is simple if no vertical line; determinant works always).
- Compute slopes or determinant using the coordinates.
- If slopes are equal (or determinant = 0), points are collinear; otherwise they are not.
Why it works: Equal slopes mean the direction between pairs of points is identical, so all points lie on a single line. The determinant formula computes twice the signed area of triangle ABC; area zero means no triangle is formed, i.e., points are on a line.
- Example 1 (simple): A(1,2), B(2,4), C(3,6). Slopes: AB = (4-2)/(2-1) = 2, AC = (6-2)/(3-1) = 2. Since slopes equal, points are collinear (they lie on y = 2x).
- Example 2 (determinant): A(0,0), B(1,1), C(2,3). Determinant = 0*(1-3) + 1*(3-0) + 2*(0-1) = 0 + 3 - 2 = 1 ≠ 0. So not collinear.
- Example 3 (vertical line): A(2,1), B(2,3), C(2,7). All x = 2, so points are collinear on the vertical line x = 2.
- Example 4 (coincident points): A(1,2), B(1,2), C(3,4). A and B coincide; the line through A and C contains B, so all three are collinear.
- Example 5 (area check): A(1,0), B(2,3), C(4,6). Compute x1(y2-y3)+x2(y3-y1)+x3(y1-y2)=1(3-6)+2(6-0)+4(0-3)= -3+12-12 = -3 ≠ 0, so not collinear.
- Slope equality: (y2 - y1)/(x2 - x1) = (y3 - y1)/(x3 - x1), provided denominators ≠ 0 (vertical special case).
- Determinant (collinearity condition): |x1 y1 1| |x2 y2 1| |x3 y3 1| = 0, i.e. x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2) = 0.
- Area of triangle ABC: Area = (1/2) * |x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2)|. Collinear ⇔ Area = 0.
- Equation check: Line through (x1,y1) and (x2,y2): (y - y1)(x2 - x1) = (x - x1)(y2 - y1). Plug (x3,y3) to check.
Applications and Problem Solving
Overview: In Coordinate Geometry (Class 10), we represent geometric objects using ordered pairs (x, y) on the Cartesian plane and use algebraic formulas to solve geometric problems. Key tools are distance, midpoint/section formulas, slope, equation of a line, and area of polygon (triangle) using coordinates. These let us compute lengths, locate points dividing segments in a given ratio, test collinearity, find equations of lines, and solve real-life layout and navigation problems.
Problem-solving approach:
- Sketch the points and shapes to visualise the problem.
- Assign or read coordinates of known points; label unknowns as variables if needed.
- Choose the right formula: distance, slope, section/midpoint, two-point form, or area (determinant) method.
- Compute algebraically, simplify, and interpret the result in the geometric context.
Typical applications: determining shortest distances (roads, flight routes approximations), dividing land or segments in a given ratio (surveying, property plots), checking alignment or straightness (construction, robotics), computing area from coordinates (land area estimation), plotting and detecting intersections (navigation, computer graphics).
Tips: Always draw a quick diagram, check units, and verify special cases (horizontal/vertical lines give undefined/zero slopes). Use determinant formula to avoid sign mistakes when calculating area or testing collinearity.
- 1) Section formula (internal): Find P that divides AB in ratio 2:3 internally where A(2,1) and B(8,6). Solution: x = (2·8 + 3·2)/(2+3) = 22/5 = 4.4, y = (2·6 + 3·1)/5 = 15/5 = 3. So P(4.4, 3).
- 2) Distance formula: Find distance between (1,2) and (5,6). Solution: d = √[(5-1)^2 + (6-2)^2] = √(16+16) = √32 = 4√2 units.
- 3) Equation of a line through two points: Through (1,2) and (3,8). Slope m = (8-2)/(3-1) = 3. Equation: y - 2 = 3(x - 1) → y = 3x - 1.
- 4) Area of triangle using coordinates: For triangle with vertices (0,0), (4,0), (0,3) area = 1/2·base·height = 1/2·4·3 = 6 square units. (Using determinant also gives 6.)
- 5) Collinearity test: Are A(1,2), B(2,4), C(3,6) collinear? Slopes AB = (4-2)/(2-1) = 2, BC = (6-4)/(3-2) = 2 → equal slopes → points are collinear.
- Distance between (x1,y1) and (x2,y2): d = √[(x2 - x1)^2 + (y2 - y1)^2]
- Midpoint of AB: ((x1 + x2)/2, (y1 + y2)/2)
- Section formula (internal) for point dividing AB in ratio m:n (from A to B): ((m x2 + n x1)/(m+n), (m y2 + n y1)/(m+n))
- Section formula (external) for ratio m:n: ((m x2 - n x1)/(m - n), (m y2 - n y1)/(m - n))
- Slope of line through (x1,y1) and (x2,y2): m = (y2 - y1)/(x2 - x1) (vertical line → slope undefined)
- Two-point form of line: (y - y1) = [(y2 - y1)/(x2 - x1)](x - x1)
Key Concepts
- Coordinate Plane
- A plane formed by two perpendicular number lines (x-axis and y-axis) used to locate points by ordered pairs (x,y).
- Ordered Pair (Coordinates)
- An ordered pair (x,y) gives the coordinates of a point: x is the abscissa (horizontal) and y is the ordinate (vertical).
- Origin
- The point (0,0) where the x-axis and y-axis intersect.
- Axes
- The x-axis is the horizontal number line and the y-axis is the vertical number line that form the coordinate plane.
- Quadrants
- The four regions of the coordinate plane numbered I to IV counterclockwise, starting with x>0,y>0 as Quadrant I.
- Distance Formula
- Distance between (x1,y1) and (x2,y2) is sqrt((x2−x1)^2 + (y2−y1)^2).
- Midpoint Formula
- Midpoint of segment joining (x1,y1) and (x2,y2) is ((x1+x2)/2, (y1+y2)/2).
- Section Formula (Internal Division)
- Point dividing segment joining A(x1,y1) and B(x2,y2) internally in ratio m:n has coordinates ((m x2 + n x1)/(m+n), (m y2 + n y1)/(m+n)).
- Section Formula (External Division)
- Point dividing A(x1,y1) and B(x2,y2) externally in ratio m:n has coordinates ((m x2 − n x1)/(m−n), (m y2 − n y1)/(m−n)).
- Slope (Gradient)
- Slope m of line through (x1,y1) and (x2,y2) is (y2−y1)/(x2−x1); it measures the steepness and sign indicates rise/fall.
- Equation of a Line (General Form)
- General form ax + by + c = 0 represents a straight line, where a and b are not both zero.
- Slope-Intercept Form
- y = mx + c, where m is the slope and c is the y-intercept (point where line meets y-axis).
- Point-Slope Form
- y − y1 = m(x − x1) is the equation of the line with slope m passing through (x1,y1).
- Intercept Form
- x/a + y/b = 1 is the equation of a line with x-intercept a and y-intercept b (a,b ≠ 0).
- Parallel Lines
- Two lines are parallel if they have equal slopes (m1 = m2) and different intercepts (so they do not coincide).
- Perpendicular Lines
- Two non-vertical lines are perpendicular if the product of their slopes is −1 (m1·m2 = −1).
- Collinear Points
- Points are collinear if they lie on the same straight line; equivalently, slopes between pairs are equal or the area of triangle formed is zero.
- Area of Triangle (Coordinate Method)
- Area of triangle with vertices (x1,y1),(x2,y2),(x3,y3) is (1/2)|x1(y2−y3)+x2(y3−y1)+x3(y1−y2)|.
- Perpendicular Bisector
- The line that is perpendicular to a segment at its midpoint; locus of points equidistant from the segment's endpoints.
- Inclination of a Line
- The angle θ a line makes with the positive x-axis measured counterclockwise; slope m = tan θ (θ in (−90°,90°) for finite m).
End-of-Chapter Trial Paper & Test Questions
Topic-wise questions to test your understanding of every concept in this chapter.
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State the distance formula between two points and the principle it is based on. / दो बिंदुओं के बीच दूरी सूत्र और जिस सिद्धांत पर यह आधारित है, उसे लिखिए।
Show answer
The distance between A(x1, y1) and B(x2, y2) is √[(x2 − x1)^2 + (y2 − y1)^2], derived from the Pythagorean theorem applied to the right triangle formed by the horizontal and vertical projections. / बिंदु A(x1, y1) और B(x2, y2) के बीच दूरी √[(x2 − x1)^2 + (y2 − y1)^2] है, जो क्षैतिज और ऊर्ध्वाधर प्रक्षेपों से बने समकोण त्रिभुज पर पाइथागोरस प्रमेय लगाने से व्युत्पन्न होती है।
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Find the distance between the points A(−2, 3) and B(3, −1). / बिंदु A(−2, 3) और B(3, −1) के बीच की दूरी ज्ञात कीजिए।
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AB = √[(3 − (−2))^2 + (−1 − 3)^2] = √[5^2 + (−4)^2] = √(25 + 16) = √41 units. / AB = √[(3 − (−2))^2 + (−1 − 3)^2] = √[5^2 + (−4)^2] = √(25 + 16) = √41 इकाई।
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Using the section formula, find the point dividing A(2, 3) and B(8, 7) internally in the ratio 3:2. / अनुभाग सूत्र का प्रयोग करके A(2, 3) और B(8, 7) को 3:2 अनुपात में आंतरिक रूप से विभाजित करने वाला बिंदु ज्ञात कीजिए।
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P = ((n·x1 + m·x2)/(m+n), (n·y1 + m·y2)/(m+n)) = ((2·2 + 3·8)/5, (2·3 + 3·7)/5) = (28/5, 27/5) = (5.6, 5.4). / P = ((n·x1 + m·x2)/(m+n), (n·y1 + m·y2)/(m+n)) = ((2·2 + 3·8)/5, (2·3 + 3·7)/5) = (28/5, 27/5) = (5.6, 5.4)।
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Find the midpoint of the segment joining (−4, 6) and (2, −2). / (−4, 6) और (2, −2) को मिलाने वाले खंड का मध्यबिंदु ज्ञात कीजिए।
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Midpoint M = ((−4 + 2)/2, (6 + (−2))/2) = (−1, 2). / मध्यबिंदु M = ((−4 + 2)/2, (6 + (−2))/2) = (−1, 2)।
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Find the area of the triangle with vertices A(1, 2), B(4, 6), C(5, 2). / शीर्ष A(1, 2), B(4, 6), C(5, 2) वाले त्रिभुज का क्षेत्रफल ज्ञात कीजिए।
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Area = (1/2)|x1(y2 − y3) + x2(y3 − y1) + x3(y1 − y2)| = (1/2)|1(6 − 2) + 4(2 − 2) + 5(2 − 6)| = (1/2)|4 + 0 − 20| = 8 square units. / क्षेत्रफल = (1/2)|x1(y2 − y3) + x2(y3 − y1) + x3(y1 − y2)| = (1/2)|1(6 − 2) + 4(2 − 2) + 5(2 − 6)| = (1/2)|4 + 0 − 20| = 8 वर्ग इकाई।
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Show that the points A(0, 0), B(2, 2), C(4, 4) are collinear. / दर्शाइए कि बिंदु A(0, 0), B(2, 2), C(4, 4) संरेख हैं।
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Computing the area: (1/2)|0(2 − 4) + 2(4 − 0) + 4(0 − 2)| = (1/2)|0 + 8 − 8| = 0; since the area of the triangle is zero, the three points are collinear. / क्षेत्रफल की गणना: (1/2)|0(2 − 4) + 2(4 − 0) + 4(0 − 2)| = (1/2)|0 + 8 − 8| = 0; चूँकि त्रिभुज का क्षेत्रफल शून्य है, तीनों बिंदु संरेख हैं।
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Why does a zero value of the coordinate area expression indicate collinear points? / निर्देशांक क्षेत्रफल व्यंजक का शून्य मान संरेख बिंदुओं को क्यों दर्शाता है?
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The expression equals twice the signed area of the triangle formed by the three points; if this area is zero, no genuine triangle is formed, meaning all three points lie on one straight line. / यह व्यंजक तीन बिंदुओं से बने त्रिभुज के चिह्नित क्षेत्रफल का दोगुना होता है; यदि यह क्षेत्रफल शून्य है, तो कोई वास्तविक त्रिभुज नहीं बनता, अर्थात् तीनों बिंदु एक सरल रेखा पर स्थित हैं।
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Find the slope of the line through (1, 2) and (3, 8) and write its equation in point-slope form. / (1, 2) और (3, 8) से होकर जाने वाली रेखा का ढाल ज्ञात कीजिए और इसका समीकरण बिंदु-ढाल रूप में लिखिए।
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Slope m = (8 − 2)/(3 − 1) = 3; using point-slope form y − y1 = m(x − x1), the equation is y − 2 = 3(x − 1), i.e., y = 3x − 1. / ढाल m = (8 − 2)/(3 − 1) = 3; बिंदु-ढाल रूप y − y1 = m(x − x1) का प्रयोग करते हुए समीकरण y − 2 = 3(x − 1), अर्थात् y = 3x − 1 है।
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