Please answer 9 and 11
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(at², 2at ) is a point lies on the line 2x -5y +12 a =0
so , put x = at² and y = 2at
2(at²) -5(2at) +12a = 0
2at² -10at +12a = 0
2a{ t² -5t +6} = 0
(t -2)(t -3) = 0
t = 2, 3
so, points are { a(2)², 2a(2)} , { a(3)² , 2a(3)}
( 4a , 4a ) and (9a, 6a) are two points
11)
(a) let slope of line is m
then equation of line is
y-3 = m( x -4)
y = mx -4m +3 --------(1)
equation of line joining the points (4, 1) and (2, 5) is
y -1 = (5-1)/(2-4) ( x -4)
y -1 = -2(x -4)
y -1 = -2x +8
y+2x =9
intersecting point P is both lines
9 -2x = mx -4m +3
6 = ( m +2)x -4m
x = (6+4m)/(m +2)
and y = 9 -(12+8m)/(m+2)
=(9m +18 -12 -8m)/(m+2)
=(6+m)/(m+2)
then,
AP/PB = 3/4
16AP² = 9PB²
16{ (4-(6 +4m)/(m+2)}² +{(1-(6+m)/(m+2)}² } = 9{ 2 -( 6+4m)/(m+2))² +(5-(6+m)/(m+2)}²
solve it then ,
m = 1/3 and -7/3
so, equation of lines are
3y -x = 5
and 3y +7x = 37
.
(b) in the same way you solve this ,
so , put x = at² and y = 2at
2(at²) -5(2at) +12a = 0
2at² -10at +12a = 0
2a{ t² -5t +6} = 0
(t -2)(t -3) = 0
t = 2, 3
so, points are { a(2)², 2a(2)} , { a(3)² , 2a(3)}
( 4a , 4a ) and (9a, 6a) are two points
11)
(a) let slope of line is m
then equation of line is
y-3 = m( x -4)
y = mx -4m +3 --------(1)
equation of line joining the points (4, 1) and (2, 5) is
y -1 = (5-1)/(2-4) ( x -4)
y -1 = -2(x -4)
y -1 = -2x +8
y+2x =9
intersecting point P is both lines
9 -2x = mx -4m +3
6 = ( m +2)x -4m
x = (6+4m)/(m +2)
and y = 9 -(12+8m)/(m+2)
=(9m +18 -12 -8m)/(m+2)
=(6+m)/(m+2)
then,
AP/PB = 3/4
16AP² = 9PB²
16{ (4-(6 +4m)/(m+2)}² +{(1-(6+m)/(m+2)}² } = 9{ 2 -( 6+4m)/(m+2))² +(5-(6+m)/(m+2)}²
solve it then ,
m = 1/3 and -7/3
so, equation of lines are
3y -x = 5
and 3y +7x = 37
.
(b) in the same way you solve this ,
Ananya220200:
The answer for the 11.a is wrong...
Answered by
1
Given equation of the line,
2x-5y+12a = 0
The point is (at^2,2at)
The point will surely satisfy the given equation.
So we shall take X = at^2 and Y = 2at
Now,
2at^2 - 5× 2at + 12a=0
2at^2 -10at + 12a=0
at^2 - 5at + 6a=0
at^2 - 3at -2at + 6a=0
at(t- 3) - 2a (t -3)=0
(t - 3) (at -2a)=0
t -3 = 0
t = 3
and ,
at -2a =0
t = 2
Now the points can be =( a×2^2, 2a×2 )and (a3^2 , 2a×3 )or (4a, 4a) and (9a , 6a)
11) b)
The segment in the ratio m:n are ( 7m - 9n /m+n ) , (9m + 5n / m+n)
The line is , 3x + 4y = 21
3 × 7m -9n /m+n + 4 × 9m + 5n / m+n =21
or, 21 m - 27n + 36 m + 20 n / m+n = 21
or, 21 m - 27n - 36m + 20n = 21m + 21n
or, - 28 n = -36 m
or, 7/9 =m/n
The ratio is = 7:9
11)a.
The point X2 = 3× 4 + 4×2 /3+4 = 20/7
The point Y2 = 3 × 1 + 4×5 /3+4 = 23/7
With the help of Two point form,
y - y1 / x - x1 = y1 - y2 / x1 - x2
or, y- 3 / x - 4 = 3 - 23/7 / 4 - 20/7
or, y - 3 / x-4 = -2/7 / -6/7
or, 3(y- 3 ) = x - 4
or , x - 3y + 5 = 0
or, x - 3y + 5 =0
The equation is x - 3y + 5 = 0
Hey Aru, Ami sure na...Kintu answer ta to atai, to mone hoe thik ache...
Ar questions Bengali te de T_T
2x-5y+12a = 0
The point is (at^2,2at)
The point will surely satisfy the given equation.
So we shall take X = at^2 and Y = 2at
Now,
2at^2 - 5× 2at + 12a=0
2at^2 -10at + 12a=0
at^2 - 5at + 6a=0
at^2 - 3at -2at + 6a=0
at(t- 3) - 2a (t -3)=0
(t - 3) (at -2a)=0
t -3 = 0
t = 3
and ,
at -2a =0
t = 2
Now the points can be =( a×2^2, 2a×2 )and (a3^2 , 2a×3 )or (4a, 4a) and (9a , 6a)
11) b)
The segment in the ratio m:n are ( 7m - 9n /m+n ) , (9m + 5n / m+n)
The line is , 3x + 4y = 21
3 × 7m -9n /m+n + 4 × 9m + 5n / m+n =21
or, 21 m - 27n + 36 m + 20 n / m+n = 21
or, 21 m - 27n - 36m + 20n = 21m + 21n
or, - 28 n = -36 m
or, 7/9 =m/n
The ratio is = 7:9
11)a.
The point X2 = 3× 4 + 4×2 /3+4 = 20/7
The point Y2 = 3 × 1 + 4×5 /3+4 = 23/7
With the help of Two point form,
y - y1 / x - x1 = y1 - y2 / x1 - x2
or, y- 3 / x - 4 = 3 - 23/7 / 4 - 20/7
or, y - 3 / x-4 = -2/7 / -6/7
or, 3(y- 3 ) = x - 4
or , x - 3y + 5 = 0
or, x - 3y + 5 =0
The equation is x - 3y + 5 = 0
Hey Aru, Ami sure na...Kintu answer ta to atai, to mone hoe thik ache...
Ar questions Bengali te de T_T
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