Question

16. Find the coordinates of the foot of the perpendicular from the origin on the straight line 3x+2y = 13.

Answer 1

Solution:-

given by:- the coordinates of the foot of the perpendicular from the origin on the straight line 3x+2y = 13.

we have:

according to question we have to find the point P .
The line OP is perpendicular to the 3 x + 2 y = 13.

》Equation of the line OP

》2 x - 3 y + k = 0

The line OP is passing through the origin (0 , 0)

》2× (0) - 3× (0) + k = 0

》k = 0

Therefore the equation of the line OP

》2 x - 3y + 0 = 0

》2 x - 3 y = 0

Both lines are intersecting at the point P.

》3 x + 2 y = 13  ............. (1)

》2 x - 3 y = 0   ...............(2)

》(1) x 3 =   9 x + 6 y = 39

》(2) x 2 =   4 x - 6 y = 0

               .....................

》                 13 x = 39

》                  x = 39/13

》                  x = 3

Substitute x = 3 in the second equation

》     2 (3) - 3 y = 0

》      6 - 3 y = 0

》    - 3 y = -6

》      y = (-6)/(-3)

》       y = 2


Therefore the required point P is (3 , 2)

Answer 2

Given equation of a line

3x + 2y = 13 ----( 1 )

i ) slope of a line ( m1 ) = -a/b = -3/2

slope of a line perpendicular to

3x + 2y = 13 is ( m2) = -1/m1

=> m2 = 2/3

ii ) Equation of a line whose slope

m2= 2/3 , and passing through

origin O( 0,0 ) is

y =mx

=> y = (2/3)x ---( 2 )

iii ) Intersecting point of ( 1 ) and (2)

is M ,

substitute y = (2/3 )x in ( 1 ),

3x + 2 ×(2/3)x = 13

=> 3x +4x/3 = 13

=> ( 9x+4x)/3=13

=> 13x/3=13

x = 39/13 = 3

substitute x = 3 in ( 2 ), we get

y = (2/3 ) × ( 3)

y = 2

Required point( M ) = ( 3 ,2)

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