{"slug":"cbse-class-10-maths-coordinate-geometry","title":"Coordinate Geometry","description":"The distance formula, the section formula and the mid-point formula, weighted towards the equidistant-point, ratio-of-division and mid-point questions set across the 2026 papers.","board":"CBSE","grade":"Class 10","subject":"Maths","chapter":"Coordinate Geometry","url":"https://www.flipnlearn.app/deck/cbse-class-10-maths-coordinate-geometry","studyUrl":"https://www.flipnlearn.app/deck/cbse-class-10-maths-coordinate-geometry/study","cards":[{"id":"abscissa","kind":"term-definition","front":"Abscissa","back":"The x-coordinate of a point, equal to its distance from the y-axis. On the x-axis the ordinate is zero, so every point there has the form (x, 0)."},{"id":"ordinate","kind":"term-definition","front":"Ordinate","back":"The y-coordinate of a point, equal to its distance from the x-axis. On the y-axis the abscissa is zero, so every point there has the form (0, y)."},{"id":"distance-formula","kind":"formula","front":"Distance formula","back":"The distance between P(x₁, y₁) and Q(x₂, y₂) is PQ = √[(x₂ − x₁)² + (y₂ − y₁)²]. It comes from applying Pythagoras theorem to the right triangle whose legs are the horizontal and vertical gaps between the points."},{"id":"distance-from-the-origin","kind":"formula","front":"Distance from the origin","back":"The distance of P(x, y) from O(0, 0) is OP = √(x² + y²), the distance formula with one point at the origin."},{"id":"why-the-order-of-points-does-not-matter","kind":"concept-question","front":"Why does it not matter whether you use (x₂ − x₁) or (x₁ − x₂) in the distance formula, and why is only the positive root taken?","back":"Each difference is squared, and a number and its negative have the same square, so both orders give the same value. The positive square root is taken because a distance can never be negative."},{"id":"how-to-classify-a-triangle-from-its-vertices","kind":"worked-step","front":"How do you check whether (3, 2), (−2, −3) and (2, 3) form a triangle and name its type?","back":"Find all three distances: PQ = √50, QR = √52 and PR = √2. Any two add to more than the third, so the points form a triangle. Since PQ² + PR² = QR², the converse of Pythagoras theorem gives a right angle at P."},{"id":"how-to-show-four-points-form-a-square","kind":"worked-step","front":"How do you use the distance formula to show that four points are the vertices of a square?","back":"Find all four sides and both diagonals. In the chapter's example each side is √34 and each diagonal √68; equal sides with equal diagonals make a square. Alternatively, four equal sides with AD² + DC² = AC² give a right angle, which also proves it."},{"id":"collinear-points","kind":"term-definition","front":"Collinear points","back":"Points that lie on one straight line. With the distance formula, three points are collinear when two of the distances add up to the third, as with AB = 3√2, BC = 2√2 and AC = 5√2 in the chapter's classroom example."},{"id":"how-to-find-the-relation-for-an-equidistant-point","kind":"worked-step","front":"How do you find a relation between x and y so that P(x, y) is equidistant from (7, 1) and (3, 5)?","back":"Set AP² = BP², giving (x − 7)² + (y − 1)² = (x − 3)² + (y − 5)², and expand. The x² and y² terms cancel, leaving x − y = 2. This line is the perpendicular bisector of AB, where every equidistant point must lie."},{"id":"how-to-find-an-equidistant-point-on-an-axis","kind":"worked-step","front":"How do you find the point on the y-axis that is equidistant from A(6, 5) and B(−4, 3)?","back":"Any point on the y-axis is (0, y), so write AP² = BP²: 6² + (5 − y)² = (−4)² + (3 − y)². The y² terms cancel, giving 4y = 36 and y = 9, so the point is (0, 9). Check that both distances come to √52."},{"id":"section-formula","kind":"formula","front":"Section formula","back":"The point dividing the segment from A(x₁, y₁) to B(x₂, y₂) internally in the ratio m₁ : m₂ is ((m₁x₂ + m₂x₁)/(m₁ + m₂), (m₁y₂ + m₂y₁)/(m₁ + m₂)). Here PA : PB = m₁ : m₂."},{"id":"section-formula-for-ratio-k-to-1","kind":"formula","front":"Section formula for a ratio k : 1","back":"The point dividing the join of A(x₁, y₁) and B(x₂, y₂) in the ratio k : 1 is ((kx₂ + x₁)/(k + 1), (ky₂ + y₁)/(k + 1)). Using one unknown k makes it easy to find an unknown ratio."},{"id":"midpoint-formula","kind":"formula","front":"Midpoint formula","back":"The mid-point of the join of A(x₁, y₁) and B(x₂, y₂) is ((x₁ + x₂)/2, (y₁ + y₂)/2). It is the section formula with the ratio 1 : 1."},{"id":"how-to-find-the-ratio-in-which-a-point-divides-a-segment","kind":"worked-step","front":"How do you find the ratio in which (−4, 6) divides the segment joining A(−6, 10) and B(3, −8)?","back":"Let the ratio be k : 1 and equate the x-coordinate from the section formula: −4 = (3k − 6)/(k + 1). This gives 7k = 2, so the ratio is 2 : 7. Then check that the y-coordinate also comes out as 6."},{"id":"how-to-find-where-an-axis-cuts-a-segment","kind":"worked-step","front":"How do you find the ratio in which the y-axis divides the segment joining (5, −6) and (−1, −4)?","back":"Take the ratio as k : 1; the dividing point's x-coordinate is (−k + 5)/(k + 1). Any point on the y-axis has abscissa 0, so −k + 5 = 0, k = 5 and the ratio is 5 : 1. Putting k = 5 into the y-coordinate gives the point (0, −13/3)."},{"id":"how-to-find-points-of-trisection","kind":"worked-step","front":"How do you find the points of trisection of the segment joining A(2, −2) and B(−7, 4)?","back":"The trisection points P and Q split AB into three equal parts, so P divides it in the ratio 1 : 2 and Q in the ratio 2 : 1. The section formula then gives P(−1, 0) and Q(−4, 2); Q is also the mid-point of PB."},{"id":"how-to-find-a-missing-vertex-of-a-parallelogram","kind":"worked-step","front":"If A(6, 1), B(8, 2), C(9, 4) and D(p, 3) are vertices of a parallelogram in order, how do you find p?","back":"The diagonals of a parallelogram bisect each other, so AC and BD share a mid-point. Equate the x-coordinates of the two mid-points: (6 + 9)/2 = (8 + p)/2, giving p = 7."},{"id":"internal-and-external-division","kind":"concept-question","front":"What is the difference between a point dividing a segment internally and externally?","back":"Internal division means the point lies between A and B on the segment, which is the case the section formula here covers. If the point lies on line AB but outside the segment, it divides AB externally; that case is left for higher classes."}]}