1708\hspace{-16}$Here $y=\int_{0}^{x}2.\mid t\mid dt\;,$ where $0<t<x$\\\\ So Using $\mathbf{Leibnitz Theorem}$\\\\ $\frac{dy}{dx}=2.\mid x\mid =2x\;\;, \left(0<t<x\right)$\\\\ So $m=\frac{dy}{dx}|_{\left(x_{1},y_{1}\right)}=2x_{1}=1\Leftrightarrow x_{1}=\frac{1}{2}$\\\\ So $y_{1}=\int_{0}^{\frac{1}{2}}2.tdt=\left[t^2\right]|_{0}^{\frac{1}{2}}=\frac{1}{4}$\\\\ So Equation of Tangent is \\\\ $y-\frac{1}{4}=1.\left(x-\frac{1}{2}\right)$\\\\ So $x-y=\frac{1}{4}$\\\\