Question

if the lines 4x+3y-1=0, x-y+5=0 and kx+5y-3=0 are concurrent then k=​

Answer 1

Step-by-step explanation:

4x+3y-1=0

4x+3y=1........(i)

x-y+5=0

x-y=-5...........(ii)

equation (ii) multiply by 3

3x-3y=-15........(iii)

Adding equation (i)&(iii)

4x+3y=1

+

3x-3y=-15

7x=-15

x=-15/7

x=-2.143

x=-2.143 put on equation (ii)

x-y=-5

-2.143-y=-5

-y=-5+2.143

-y=-2.852

y=2.852

put the value x and y in equation

kx+5y-3=0

k(-2.143)+5(2.852)-3=0

-2.143k+14.26-3=0

-2.143k+11.26=0

-2.143k=-11.26

k=-11.26+2.143

k=-9.117

k= -9.117