For x = 2, y = 1, giving (2, 1) in Quadrant I. For x = -1, y = -4, giving (-1, -4) in Quadrant III. Both points satisfy 5x – 3y = 7. Class 9 Maths Ganita Manjari Part 2 ...
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For x = 2, y = -1, giving the point (2, -1) in Quadrant IV. For x = -1, y = 4, giving the point (-1, 4) in Quadrant II. Both satisfy the equation. Class 9 Maths Ganita Manjari Part 2 ...
Substituting (2, -1) into 2mx + 3y = 7 yields 4m – 3 = 7, which gives m = 2.5. Substituting into 4x + ny = -10 gives 8 – n = -10, resulting in n = 18. Ganita Manjari part ...
(i) By the Midpoint Theorem, PQ = BC/2 = BR = RC, QR = AB/2 = AP = PB and PR = AC/2 = AQ = QC, so △PQR ≅ △QPA ≅ △RPB ≅ △CQR by SSS. (ii) Yes, construct ...
Adjacent angles of a parallelogram sum to 180°, so their bisectors add to 90°, forcing right angles at intersections to form a rectangle. If AB = BC, all bisectors intersect at a single point. Cbse Class 9 Maths Ganita Manjari Part ...