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Question is in the image attached

Asked by swayamagarwal2114 | 15 Jul, 2022, 09:19: AM

Expert Answer

As the triangle slides

Let A(a, 0) and B(0, b) be the two vertices of triangle and C(p, q) be the third vertex.

Suppose (h, k) be the centroid of the triangle

rightwards double arrow straight a equals square root of 2 cosθ comma space straight b equals square root of 2 sinθ Now comma space BC space and space AC space are space inclined space at space left parenthesis 45 to the power of straight o minus straight theta right parenthesis space and space 180 to the power of straight o minus left parenthesis 45 to the power of straight o minus straight theta right parenthesis space straight w. straight r. straight t space straight x minus axs space respectively. rightwards double arrow straight p equals cos left parenthesis 45 minus straight theta right parenthesis space and space straight q equals square root of 2 space sinθ space plus space sin left parenthesis 45 to the power of straight o minus straight theta right parenthesis Also comma space straight p equals square root of 2 space cosθ space minus space cos left parenthesis 45 to the power of straight o plus straight theta right parenthesis space and space straight q equals sin left parenthesis 45 to the power of straight o plus straight theta right parenthesis Now comma space straight h equals fraction numerator straight p plus straight a over denominator 3 end fraction space and space straight k equals fraction numerator straight q plus straight b over denominator 3 end fraction rightwards double arrow 3 straight h equals cos left parenthesis 45 to the power of straight o minus straight theta right parenthesis plus square root of 2 space cosθ equals fraction numerator 3 over denominator square root of 2 end fraction cosθ plus fraction numerator 1 over denominator square root of 2 end fraction sinθ rightwards double arrow 3 square root of 2 straight h equals 3 cosθ plus sinθ space space... space left parenthesis 1 right parenthesis And comma space 3 straight k equals square root of 2 space sinθ space plus space sin left parenthesis 45 to the power of straight o minus straight theta right parenthesis plus square root of 2 sinθ rightwards double arrow 3 straight k equals 2 square root of 2 sinθ plus fraction numerator 1 over denominator square root of 2 end fraction cosθ minus fraction numerator 1 over denominator square root of 2 end fraction sinθ rightwards double arrow 3 square root of 2 straight k equals 3 sinθ plus cosθ space space... space left parenthesis 2 right parenthesis Solving space left parenthesis 1 right parenthesis space and space left parenthesis 2 right parenthesis comma space we space get space sinθ equals fraction numerator 3 square root of 2 over denominator 8 end fraction left parenthesis 3 straight k minus straight h right parenthesis space and space cosθ equals fraction numerator 3 square root of 2 over denominator 8 end fraction left parenthesis 3 straight h minus straight k right parenthesis Since comma space sin squared straight theta plus cos squared straight theta equals 1 rightwards double arrow fraction numerator 18 left parenthesis 3 straight k minus straight h right parenthesis squared over denominator 64 end fraction plus fraction numerator 18 left parenthesis 3 straight h minus straight k right parenthesis squared over denominator 64 end fraction equals 1 rightwards double arrow left parenthesis 3 straight x minus straight y right parenthesis squared plus left parenthesis 3 straight y minus straight x right parenthesis squared equals 32 over 9

Answered by Renu Varma | 19 Jul, 2022, 12:01: PM

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