Answer
See below
Work Step by Step
Given $p_1(x)=a+bx\\
p_2(x)=c+dx$
Obtain $\lambda_1p_1(x)+\lambda_2p_2(x)\\
=\lambda_1(a+bx)+\lambda_2(c+dx)\\
=\lambda_1a+\lambda_2 c+x(\lambda_1 b+\lambda_2 d)$
This sum is equal to zero if and only if
$\lambda_1 a+\lambda_2 c=0\\
\lambda_1 b+\lambda_2 d=0\\
\rightarrow \lambda_1=-\lambda_2\frac{c}{a}\\
\lambda_2=-\lambda_1 \frac{b}{d}=\lambda_2 \frac{c}{a}\frac{b}{d}$
Since $p_1$ and $p_2$ are linearly independent
$\lambda_2(1-\frac{cb}{ad})\ne 0 \rightarrow \lambda_2\frac{ad-cb}{ad}\ne0\\
\rightarrow \lambda_2 \ne 0$
and therefore $ad\ne cb$
