Math, asked by Anonymous, 1 year ago

Please answer 9 and 11

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Answered by abhi178
2
(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 ,


Ananya220200: The answer for the 11.a is wrong...
Ananya220200: Please do check your answer
abhi178: what ??
abhi178: it is correct okay , again read
Ananya220200: The answer is given x - 3y + 5 =0 in the answers
abhi178: okay i will checked this at night , bit busy
Ananya220200: Okay
Answered by Ananya220200
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
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