if the distance of the point(a,b) from(-3,0) as well as(3,0) are 4each,find a and b
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X(-3,0)
Y(3,0)
A(a,b)
by sectional formula:
XA = 4 = √[(-3-a)²+(0-b)²]
16 = 9 + a² + 6a + b²
7 = a² + 6a + b² ___ (i)
YA = 4 = √[(3-a)²+(0-b)²]
16 = 9 + a² - 6a + b²
7 = a² - 6a + b² ___ (ii)
from (i) and (ii)
a² + 6a - b² = a² - 6a + b²
6a = - 6a
a = -a
a + a = 0
2a = 0
therefore a = 0
putting a = 0 in (i):
0² + 6*0 + b² = 7
b² = 7
b = ±√7
therefore A is at (0,√7) or (0,-√7)
Y(3,0)
A(a,b)
by sectional formula:
XA = 4 = √[(-3-a)²+(0-b)²]
16 = 9 + a² + 6a + b²
7 = a² + 6a + b² ___ (i)
YA = 4 = √[(3-a)²+(0-b)²]
16 = 9 + a² - 6a + b²
7 = a² - 6a + b² ___ (ii)
from (i) and (ii)
a² + 6a - b² = a² - 6a + b²
6a = - 6a
a = -a
a + a = 0
2a = 0
therefore a = 0
putting a = 0 in (i):
0² + 6*0 + b² = 7
b² = 7
b = ±√7
therefore A is at (0,√7) or (0,-√7)
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