Question

The equation of a line is 4x – 3y + 4 = 0. Find the equation of a line perpendicular to the line 4x – 3y + 4 = 0 and passing through the point of intersection of the lines shown by the equations x – y + 2 = 0 and 3x + y – 10 = 0.

Answer 1

Answer: 3x + 4y - 22 = 0

Step-by-step explanation:

1) slope of the line = negative inverse of the slope of the given line,

= -3/4

2) point of intersection

adding the eq's => x - y = -2

3x + y = 10

----------------

4x = 8

x = 2

For y

=> 2 - y = -2 ( as per equation 1)

=> y = 4

Point of intersection P = (2,4)

3) equation of line

=> (y - y0) = m (x - x0)

=> (y - 4) = -3/4 (x -2)

=> 4y - 16 = 6 - 3x

=> 3x + 4y - 22 = 0

Answer 2

Solution :

i ) Finding intersecting point of the

lines x - y + 2 = 0 ----( 1 )

3x + y - 10 = 0 ----( 2 )

Add ( 1 ) & ( 2 ) , we get

4x - 8 = 0

=> 4x = 8

=> x = 8/4

=> x = 2

Put x = 2 in equation ( 1 ), we get

2 - y + 2 = 0

=> 4 - y = 0

y = 4

Therefore ,

Intersecting point ( x , y ) = ( 2 ,4 )

ii ) Find Slope of a line 4x - 3y + 4 = 0:

Compare this equation with ax+by+c=0,

a = 4 , b = -3 , c = 4

Slope ( m1 ) = - a/b

= -4/(-3)

m1 = 4/3

Let Slope of a line perpendicular to

4x - 3y + 4 = 0 is m2 .

m2 = -1/m1

= -1/(4/3)

= -3/4

iii ) Equation of a line passing through

( x1 , y1 ) = ( 2 , 4 ) and whose

slope is -3/4

y - y1 = m( x - x1 )

=> y - 4 = ( -3/4 )( x - 2 )

=> 4( y - 4 ) = -3x + 6

=> 4y - 16 + 3x - 6 = 0

=> 3x + 4y - 22 = 0

Therefore ,

Required equation is

3x + 4y - 22 = 0

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