Locus

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Step 1: The condition is PA = PB. Step 2: Mark the mid-point M of AB. Step 3: The set of all such points lies on the perpendicular bisector of AB. Answer: The locus is the perpendicular bisector of AB…

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Step 1: Take any point P such that PA = PB. Step 2: Join PA, PB, and AB. Let M be the mid-point of AB. Step 3: In triangles AMP and BMP, AM = BM, MP is common, and AP = BP. Step 4: So, triangles are c…

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Step 1: Let M be the mid-point of AB. Step 2: Join QA and QB. Step 3: In triangles AMQ and BMQ, AM = BM, MQ is common, and ∠AMQ = ∠BMQ = 90 degrees. Step 4: So, triangles are congruent by RHS. Step 5:…

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Step 1: Let the fixed point be the centre. Step 2: All points at the same fixed distance from it form a circle. Step 3: The fixed distance becomes the radius. Answer: The locus is a circle with the fi…

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Use it when the condition says a point must remain at the same distance from one fixed point. Step-by-step idea: 1. The centre stays fixed. 2. The distance stays unchanged. 3. All such points make a c…

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Step 1: Two intersecting lines form four angles. Step 2: Points equally distant from both lines lie on the angle bisectors. Step 3: There are two bisector lines, one for each pair of vertically opposi…

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Step 1: Two intersecting lines create four angles. Step 2: A point equally distant from both lines can lie in any one of the angle regions. Step 3: The two angle bisectors cover all such positions. St…

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Step 1: Given PM = PN. Step 2: Draw OP where O is the intersection of AB and CD. Step 3: In right triangles OPM and OPN, OP is common, and ∠OMP = ∠ONP = 90 degrees. Step 4: So, triangles OPM and OPN a…