Question

Show that the lines x-y=6, 4x-3y=20 and 6x+5y+8=0 are concurrent . Also find the
point of concurrence
​

Answer 1

We know that two or more lines are said to be concurrent if they intersect at a single point.

We first find the point of intersection of the first two lines.

2x + 5y = 1 ....(1)

x - 3y = 6 ....(2)

Multiplying (2) by 2, we get,

2x − 6y = 12 ....(3)

Subtracting (3) from (1), we get,

11y = −11

y = −1

From (2), x = 6 + 3y = 6 − 3 = 3

So, the point of intersection of the first two lines is (3, −1).

If this point lie on the third line, i.e., x + 5y + 2 = 0, then the given lines will be concurrent. Substituting x = 3 and y = -1, we have:

L.H.S. = x + 5y + 2

= 3 + 5(-1) + 2

= 5 - 5

= 0 = R.H.S.

Thus, (3, −1) also lie on the third line.

Hence, the given lines are concurrent.

Hence, the given line are concurrent and their point of intersection is P(2,-4).

Answer 2

Step-by-step explanation:

On solving x−y−6=0 and 4x−3y−20=0

we get

x=2,y=−4

Also,

x=2,y=−4 satisfies 6x+5y+8=0 as (6×2)+[5×(−4)]+8=0

point of intersection of the given lines is (2,−4)