NCERT Solutions · Class 10 · Chapter 7: Coordinate Geometry
NCERT Solutions for Class 10 Maths Chapter 7 Exercise 7.1
Exercise 7.1: Distance formula. \(PQ = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\). Use it to test collinearity (the two shorter distances add to the longest), classify triangles and quadrilaterals (compare sides and diagonals) and find equidistant points.
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Without a calculator, square: \[(\sqrt5 + \sqrt{212})^2 = 217 + 2\sqrt{1060}\], which equals \(265\) only if \(\sqrt{1060} = 24\); but \(24^2 = 576 \ne 1060\). So \(AB + BC \ne AC\), and no other pair adds up either (\(\sqrt{265}\) is the longest side).
Name the quadrilateral formed (if any) by the points, and give a reason.
(i) \((-1, -2),\ (1, 0),\ (-1, 2),\ (-3, 0)\)
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All sides \(\sqrt8 = 2\sqrt2\); both diagonals \(4\).
Answer: Square
(ii) \((-3, 5),\ (3, 1),\ (0, 3),\ (-1, -4)\)
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\((-3, 5), (0, 3), (3, 1)\): distances \(\sqrt{13} + \sqrt{13} = 2\sqrt{13}\), which is the distance from \((-3, 5)\) to \((3, 1)\). Three of the points are collinear.
Answer: No quadrilateral (three points are collinear)
(iii) \((4, 5),\ (7, 6),\ (4, 3),\ (1, 2)\)
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Opposite sides: \(\sqrt{10}, \sqrt{10}\) and \(\sqrt{13}, \sqrt{13}\).
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