## Trigonometry 7th Edition

To calculate VALUE OF 'x' , we will use distance formula- r= $\sqrt { (x_{2} - x_{1}) ^{2} + (y_{2} - y_{1}) ^{2}}$ As per given data $\sqrt 13$ = $\sqrt { (1 - x) ^{2} + (5 -2) ^{2}}$ $\sqrt 13$ = $\sqrt { 1 - 2x +x^{2} + 9 }$ [ on expanding $(1 - x)^{2}$] $\sqrt 13$ = $\sqrt { x^{2} - 2x + 10 }$ Squaring both the sides- 13 = $x^{2} - 2x + 10$ Rearranging- $x^{2} - 2x -3$ = 0 Factorizing the L.H.S.- (x+1) (x-3) = 0 Therefore Either x = -1 or 3