# Which is the intercepting point of the lines -4x+16y-20 = 0 and 12x-y-1 = 0?

hala718 | Certified Educator

d1: -4x + 16y - 20 = 0

d2: 12x - y - 1 = 0

Theintercepting point is where d1 = d2

Then,

d1: y = (1/4)x  + 5/4

d2: y= 12x -1

==> (1/4)x + 5/4 = 12x - 1

==> 12x - (1/4x) = (5/4) + 1

==> (47/4)x = 9/4

==> x = (9/4)*(4/47) = 9/47

==> X = 9/47

==> Y= 12X -1 = 12(9/47) - 1

= 108/47 - 1

= 61/47

Then, both lines intersect at the point ( 9/47, 61/47)

neela | Student

To find the intercepting point of -4x+16y-20 = 0 ........(1)and

12x-y-1 = 0...........(2), we solve the equations and find the x and y coordinte values .

3 * Eq (1) + eq (2) eliminates x:

3(-4x +16y-20) +12x-y-1 = 3*0+0 = 0

-12x +48y  - 60 +12x-y-1 = 0

48y-y - 60 -1 = 0

47y = 61

y = 61/47.

Fro 1st equation, -4x+16y-20 = 0, weget -4x+16(61/47)-20 = 0

-4x = 20 - 16*60/47 = (20*47- 61*16)/47 = -36/47

-4x = -36/47

x= (-36/47)/-4 = 9/47

x = 9/47 and y =  61/47

giorgiana1976 | Student

To find the intercepting point of the given lines, we'll have to solve the system formed form the equation of the lines.

-4x+16y-20 = 0

12x-y-1 = 0

We'll solve the system using the substitution method.

We'll note the equations of the system as:

-4x+16y = 20 (1)

12x-y = 1 (2)

We'll re-write (2):

y = 12x-1 (3)

We'll substitute y in (1):

-4x+16(12x-1) = 20

We'll remove the brackets:

-4x + 192x - 16 - 20 = 0

188x = 36

We'll divide by 188:

x = 36/188

x = 18/94

x = 9/47

We'll substitute x in (2):

12*9/47 - y = 1

y = 108/47 - 1

y = 61/47

The coordinates of the intercepting point is the solution of the system: {( 9/47 , 61/47)}.