(a) 2x – y +7=0
(b) 3x +2 y – 7=0
(c) 3x – y – 7 =0
(d) 3x + y – 7=0
(d) 3x + y – 7=0
Explanation:
Any point (x, y) of perpendicular bisector will be equidistant from A & B.
By, Symmetry
AP = BP
= `\sqrt{(x-4)^{2}+(y-5)^{2}}` = `\sqrt{(x-(-2)^{2}+(y-3)^{2}`
= `\sqrt{(x-4)^{2}+(y-5)^{2}}` = `\sqrt{(x + 2)^{2}+(y-3)^{2}`
By squaring both side
= `(x-4)^{2} + (y-5)^{2} = (x+2)^{2} +(y-3)^{2}`
= `x^{2}+16-8x+y^{2}+25-10y` = `x^{2}+ 4-4x+y^{2}+9-6y`
= `-8x-10y + 41` = `4x - 6y + 13`
= `-8x-10y + 41 - 4x + 6y - 13` = 0
= `-8x- 4x -10y+ 6y+ 41- 13` = 0
= 12x + 4y - 28 = 0
= 4 (3x + y - 7) = 0
= 3x + y - 7 = `0/4`
= 3x + y - 7 = 0
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