Question

3. Show that the straight lines 3x-5y+ 7 =0 and 15x +9y+4 = 0 are perpendicular

Answer 1

Solution:-

given by:-
the straight lines 3x-5y+ 7 =0 and 15x +9y+4 = 0 are perpendicular.

we have ,


》If two lines are perpendicular,then the product of their slopes will be equal to -1

》m(1) x m(2) = -1

》Slope of the first line 3x - 5 y + 7 = 0

》  Slope (m) = - coefficient of x/coefficient of y

》              m(1)  = -3/(-5)

 》                   = 3/5  

》Slope of the second line 15x + 9y + 4 = 0

》              m(2) = -15/9

》                   = -5/3

》          m(1) x m(2) = (3/5) x (-5/3) = -1

the product of those slopes is -1 .They are perpendicular



☆ i hope its help☆

Answer 2

Solution:

i ) Compare 3x - 5y + 7 = 0 with

ax + by + c = 0 ,

a = 3 , b = -5 , c = 7

slope of a line ( m1 ) = -a/b

=> m1 = - 3/( -5 )

=> m1 = 3/5 ---( 1 )

ii ) Similarly ,

slope of the line 15x + 9y + 4 = 0,

m2 = - 15/9

=> m2 = -5/3 --( 2 )

m1 × m2 = ( 3/5 ) × ( -5/3 ) = -1

Therefore ,

m1 × m2 = -1

Given lines are perpendicular to each

other.

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