question_answer

In \[\Delta ABC\], D and E axe points on the sides AB and AC respectively, such that\[DE||BC\].  If \[AD=4x-3\], \[AE=8x-7\], \[BD=3x-1\] and\[CE=5x-3\], then the value of x is

A) \[\frac{1}{2}\]

B) 4

C) 1

D) \[\frac{2}{3}\]

Correct Answer: C

Solution :

Given, in \[\Delta ABC,\]\[DE\,\,|\,\,\,|\,\,\,BC\]

\[\therefore\] By Thales theorem,

we get

\[\frac{AD}{DB}=\frac{AE}{EC}\Rightarrow \,\,\frac{4x-3}{3x-1}=\frac{8x-7}{5x-3}\]

\[\left[ \because \,\,\,AD=4x-3,\,DB=3x-1,\,AE=8x-7,\,EC=5x-3 \right]\]

\[\Rightarrow \,\,\,\,\left( 4x-3 \right)\left( 5x-3 \right)=\left( 8x-7 \right)\left( 3x-1 \right)\]

\[\Rightarrow \,\,\,20{{x}^{2}}-12x+9-15x=24{{x}^{2}}-21x\,-8x+7\]

\[\Rightarrow \,\,\,\,\,4{{x}^{2}}-2x-2=0\]

\[\Rightarrow \,\,\,\,2{{x}^{2}}-x-1=0\]

[dividing both sides by 2]

\[\Rightarrow \,\,\,\,2{{x}^{2}}-2x+x-1=0\]

[by splitting the middle term]

\[\Rightarrow \,\,\,\,2x\left( x-1 \right)+1\left( x-1 \right)=0\]

\[\Rightarrow \,\,\,\,\,\,\left( 2x+1 \right)\left( x-1 \right)=0\]

\[\Rightarrow \,\,\,\,\,\,x=-\frac{1}{2}\,\,or\,\,x=1\]

If \[x=-\frac{1}{2}\], then \[AD=4\times \left( -\frac{1}{2} \right)-3=-5<0\]

[not possible; since, length cannot be negative]

Hence, x = 1 is the required value.