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Exercise 6.3 · Q1

Q.Show that the lines 3x+2y+9=03x + 2y + 9 = 0 and 12x+8y−15=012x + 8y - 15 = 0 are parallel lines.

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For a1x+b1y+c1=0a_1x+b_1y+c_1=0 and a2x+b2y+c2=0a_2x+b_2y+c_2=0, the lines are parallel iff a1b2=a2b1a_1b_2=a_2b_1.

  • Here a1b2=3×8=24a_1b_2 = 3\times 8 = 24 and a2b1=12×2=24a_2b_1 = 12\times 2 = 24, so they are equal.

Two lines in general form a1x+b1y+c1=0a_1x+b_1y+c_1=0 and a2x+b2y+c2=0a_2x+b_2y+c_2=0 are parallel exactly when their direction ratios match, i.e. a1b2=a2b1a_1b_2=a_2b_1 (equivalently a1/a2=b1/b2a_1/a_2=b_1/b_2).

Step 1. Identify the coefficients. For 3x+2y+9=03x+2y+9=0: a1=3, b1=2, c1=9a_1=3,\ b_1=2,\ c_1=9. For 12x+8y−15=012x+8y-15=0: a2=12, b2=8, c2=−15a_2=12,\ b_2=8,\ c_2=-15.

Step 2. Apply the parallel test a1b2=a2b1a_1b_2=a_2b_1.

a1b2=3×8=24,a2b1=12×2=24.a_1b_2 = 3\times 8 = 24, \qquad a_2b_1 = 12\times 2 = 24.

Since a1b2=a2b1=24a_1b_2=a_2b_1=24, the two lines are parallel (their slopes are equal: m1=−32m_1=-\dfrac{3}{2} and m2=−128=−32m_2=-\dfrac{12}{8}=-\dfrac{3}{2}).

Step 3. Confirm the lines are not identical. Two parallel lines are the same line only if a1a2=b1b2=c1c2\dfrac{a_1}{a_2}=\dfrac{b_1}{b_2}=\dfrac{c_1}{c_2}. Here a1a2=312=14\dfrac{a_1}{a_2}=\dfrac{3}{12}=\dfrac14 and b1b2=28=14\dfrac{b_1}{b_2}=\dfrac{2}{8}=\dfrac14 agree, but c1c2=9−15=−35≠14\dfrac{c_1}{c_2}=\dfrac{9}{-15}=-\dfrac{3}{5}\ne \dfrac14. So the lines are parallel but distinct — they never meet.

✓Final answer

Since a1b2=a2b1=24a_1b_2=a_2b_1=24, the lines 3x+2y+9=03x+2y+9=0 and 12x+8y−15=012x+8y-15=0 are parallel (equal slope −3/2-3/2), and since c1/c2≠a1/a2c_1/c_2\ne a_1/a_2 they are two distinct parallel lines, not one and the same line.

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