Geometry & Figures

Depth · one concept from first principles · step 2 of 4

A second solution forces the two lines to coincide

The system of equations kx−3y=0kx-3y=0, 2x+y=kx2x+y=kx has a solution other than x=0x=0, y=0y=0. Find the value of the constant kk.

Expressions
Your answer
How to solve
Strategy Convert to Algebra — First reorganize the second equation so both have the form (…)x + (…)y = 0. Picture them as lines: both pass through the origin. Two different lines can share only one point, so a second shared point means the lines are identical, which turns into a proportion of coefficients.
1STEP 1

Move everything to the left side

Rewrite the second equation so both equations have 0 on the right.

2x + y = kx → (2-k)x + y = 0
2STEP 2

Picture the two lines

Both lines pass through the origin. Two different lines meet at most once, so another common solution means they are the same line.

3STEP 3

Make the coefficients proportional

Two equations of the form (…)x+(…)y=0 describe the same line when their x-coefficients and y-coefficients are in the same ratio.

k/(2-k) = -3/1
4STEP 4

Solve for k

Cross-multiply and collect the k terms.

k = -3(2-k) = -6 + 3k → -2k = -6 → k = 3
Answer
3
With k=3 the equations are 3x-3y=0 and 2x+y=3x, and both simplify to y=x. So x=1, y=1 (and every point with y=x) is a solution besides the origin.
Takeaway

Two lines through the same point share another point only if they are the same line.

  • Move everything to the left side
  • Picture the two lines
  • Make the coefficients proportional
  • Solve for k
Where next?
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