The extremities of the base of an isosceles triangle abc are the points a(2,0) and b(0,1). if the equation of the sides ac is x=2 then the slope of the side bc is
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20
heya....
equation of side ac is x=2,
so the point c has coordinates (2,k) and also distance between points a and c is equal to points b and c(since isosceles),
so (k=5/2)
and using slope formula slope of bc is 3/4
tysm.#gozmit
equation of side ac is x=2,
so the point c has coordinates (2,k) and also distance between points a and c is equal to points b and c(since isosceles),
so (k=5/2)
and using slope formula slope of bc is 3/4
tysm.#gozmit
neosingh:
i mean the trick to find the 'k'
yea :-)
i used distance formula. i don't know this method
hehe, ja kundan ja, jee le apni zindagi
:-D
Answered by
15
Let C(2,y)
by distance formula
CB = CA
square
CB^2 = CA^2
(2-0)^2 + (y-1)^2 = (2-2)^2 + (y-0)^2
4 + y^2 - 2y +1 = y^2
5 = 2y
y = 5/2
so C(2,5/2)
slope of the side bc is using slope formula
= (y2 - y1)/(x2-x1)
= (5/2 - 1)/(2-0)
= 3/4 ans
by distance formula
CB = CA
square
CB^2 = CA^2
(2-0)^2 + (y-1)^2 = (2-2)^2 + (y-0)^2
4 + y^2 - 2y +1 = y^2
5 = 2y
y = 5/2
so C(2,5/2)
slope of the side bc is using slope formula
= (y2 - y1)/(x2-x1)
= (5/2 - 1)/(2-0)
= 3/4 ans
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