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Exercise 4.3 · Q48

Q.Equations of pairs of opposite sides of a parallelogram are x2−7x+6=0x^2 - 7x + 6 = 0 and y2−14y+40=0y^2 - 14y + 40 = 0. Find the joint equation of its diagonals.

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x2−7x+6=0⇒x=1,6x^2-7x+6=0 \Rightarrow x=1,6; y2−14y+40=0⇒y=4,10y^2-14y+40=0 \Rightarrow y=4,10. The parallelogram (here a rectangle) has vertices (1,4),(6,4),(6,10),(1,10)(1,4),(6,4),(6,10),(1,10). Diagonal 1 through (1,4),(6,10)(1,4),(6,10): slope 65\tfrac65, equation 6x−5y+14=06x-5y+14=0. Diagonal 2 through (6,4),(1,10)(6,4),(1,10): slope −65-\tfrac65, equation 6x+5y−56=06x+5y-56=0. Combined …

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