Skip to main contentSkip to solution

The value of kk for which the following lines are concurrent is

x−y−1=02x+3y−12=02x−3y+k=0x-y-1=0 \newline 2x+3y-12=0 \newline 2x-3y+k=0

Solution

✅ Correct Option: 3

Three lines are concurrent when they all meet at the same point.


Given lines:

  • Line 1: x−y−1=0x - y - 1 = 0
  • Line 2: 2x+3y−12=02x + 3y - 12 = 0
  • Line 3: 2x−3y+k=02x - 3y + k = 0

First, find where lines 1 and 2 intersect.

From line 1: x=y+1x = y + 1

Substitute into line 2:

2(y+1)+3y−12=02(y + 1) + 3y - 12 = 0

2y+2+3y−12=02y + 2 + 3y - 12 = 0

5y−10=05y - 10 = 0

y=2y = 2

Therefore: x=y+1=2+1=3x = y + 1 = 2 + 1 = 3

Lines 1 and 2 intersect at (3,2)(3, 2)


For all three lines to be concurrent, line 3 must pass through (3,2)(3, 2).

Substitute (3,2)(3, 2) into line 3:

2x−3y+k=02x - 3y + k = 0

2(3)−3(2)+k=02(3) - 3(2) + k = 0

6−6+k=06 - 6 + k = 0

k=0k = 0


Therefore, k=0k = 0

Keyboard Shortcuts

  • Left arrow: Previous question
  • Right arrow: Next question
  • S key: Jump to solution
  • Q key: Jump to question