prove that point(7,5) is equidistant from points (2,4) and(6,10)
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Answered by
13
let points be A(7,5) B(2,4) C(6,10)
THE DISTANCE BETWEEN A AND B=√(7-2)²+(5-4)²
=√25+1=√26 UNITS
THE DISTANCE BETWEEN A AND C=√(7-6)²+(5-10)²
=√1+25=√26 UNITS.
since AB=AC..A IS EQUIDISATANT FROM B AND C
THE DISTANCE BETWEEN A AND B=√(7-2)²+(5-4)²
=√25+1=√26 UNITS
THE DISTANCE BETWEEN A AND C=√(7-6)²+(5-10)²
=√1+25=√26 UNITS.
since AB=AC..A IS EQUIDISATANT FROM B AND C
Answered by
11
A(2,4), B(6,10) P(7,5)
let us do without calculating squares and square roots.
So P lies on the perpendicular bisector of AB, if PA = PB.
Mid point C of AB is = C(4, 7).
Slope of AB = 6/4 = 3/2
Slope of PC = 2/-3 = -2/3 Product of slopes is -1.
So CP is perpendicular bisector of AB. It P is equi distant from A and B
let us do without calculating squares and square roots.
So P lies on the perpendicular bisector of AB, if PA = PB.
Mid point C of AB is = C(4, 7).
Slope of AB = 6/4 = 3/2
Slope of PC = 2/-3 = -2/3 Product of slopes is -1.
So CP is perpendicular bisector of AB. It P is equi distant from A and B
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