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  • The perpendicular bisector of the line segment joining the points A (1, 5) andB (4, 6) cuts the y-axis at (A) (0, 13) (B) (0, –13) (C) (0, 12) (D) (13, 0)

The perpendicular bisector of the line segment joining the points A (1, 5) andB (4, 6) cuts the y-axis at

(A) (0, 13)

(B) (0, –13)

(C) (0, 12)

(D) (13, 0)

Answers (1)

\\At y-axis, x = 0 \therefore point P is (0, y).\\ A (1, 5) and B (4, 6)\\ \therefore AP = BP\\

Squaring both sides we get

\\AP\textsuperscript{2} = BP\textsuperscript{2}\\ (x\textsubscript{1} - 0)\textsuperscript{2} + (y\textsubscript{1} + y)\textsuperscript{2} = (x\textsubscript{2}- 0)\textsuperscript{2} + (x\textsubscript{2} - y)\textsuperscript{2 }\\ \textsuperscript{ }(Because distance formula =\sqrt{{{({{x}_{2}}-{{x}_{1}})}^{2}}+{{({{y}_{2}}-{{y}_{1}})}^{2}}} )\\ (1 - 0)\textsuperscript{2} + (5 - y)\textsuperscript{2} = (4 - 0)\textsuperscript{2} + (6 - y)\textsuperscript{2}\\ (1)\textsuperscript{2} + (5)\textsuperscript{2} + (y)\textsuperscript{2} - 2 \times 5 \times y = (4)\textsuperscript{2} + (6)\textsuperscript{2} + (y)\textsuperscript{2} - 2 \times 6 \times y\\ \{ using (a - b)\textsuperscript{2} = a\textsuperscript{2} + b\textsuperscript{2} - 2ab \} \\ 1 + 25 + y\textsuperscript{2} - 10y = 16 + 36 + y\textsuperscript{2} - 12y\\ 26 - 10y + 12y = 52\\ 2y = 52 - 26\\ 2y = 26\\ y = 13\\

Hence, point P is (0, 13)

Therefore, option A is correct.

Posted by

infoexpert21

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