What value(s) of x will make DE || AB, in the given figure? AD = 8x

1.	What value(s) of x will make DE \|\| AB, in the given figure? AD = 8x + 9, CD = x + 3,BE = 3x + 4, CE = x.
Answer» Given : In ABC, DE // AB AD = 8x + 9, CD = x + 3, BE = 3x + 4, CE = x By Basic proportionality theorem, If DE // AB then we should have \(\frac{CD}{DA}=\frac{CE}{EB}\) \(\frac{x+3}{8x+9}\) = \(\frac{x}{3x+4}\) ⇒ (x + 3) (3x + 4) = x (8x + 9) ⇒ x (3x + 4) + 3 (3x + 4) – 8x² + 9x ⇒ 3x² + 4x + 9x + 12 = 8x² + 9x ⇒ 8x² + 9x – 3x² – 4x – 9x -12 = 0 ⇒ 5x² – 4x – 12 = 0 ⇒ 5x² – 10x + 6x – 12 = 0 ⇒ 5x (x – 2) + 6 (x – 2) = 0 ⇒ (5x + 6) (x – 2) = 0 ⇒ 5x + 6 = 0 or x – 2 = 0 ⇒ x =\(\frac{-6}{5}\) or x = 2; x cannot be negative. ∴ The value x = 2 will make DE // AB.

What value(s) of x will make DE || AB, in the given figure? AD = 8x + 9, CD = x + 3,BE = 3x + 4, CE = x.

Answer»

Given : In ABC, DE // AB AD = 8x + 9, CD = x + 3,

BE = 3x + 4, CE = x

By Basic proportionality theorem,

If DE // AB then we should have

\(\frac{CD}{DA}=\frac{CE}{EB}\)

\(\frac{x+3}{8x+9}\) = \(\frac{x}{3x+4}\)

⇒ (x + 3) (3x + 4) = x (8x + 9)

⇒ x (3x + 4) + 3 (3x + 4) – 8x² + 9x

⇒ 3x² + 4x + 9x + 12 = 8x² + 9x

⇒ 8x² + 9x – 3x² – 4x – 9x -12 = 0

⇒ 5x² – 4x – 12 = 0

⇒ 5x² – 10x + 6x – 12 = 0

⇒ 5x (x – 2) + 6 (x – 2) = 0

⇒ (5x + 6) (x – 2) = 0

⇒ 5x + 6 = 0 or x – 2 = 0

⇒ x =\(\frac{-6}{5}\) or x = 2;

x cannot be negative.

∴ The value x = 2 will make DE // AB.