Let ABC be a right angled triangle whose vertices are $A (0, 0), B (- 8, 8)$ , and $C (x, 8)$ respectively, then the possible value of x is

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Solution

The correct option is C $8$
Distance between two points $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ can be calculated
using the formula $\sqrt{{(x_{2} - x_{1})}_{2}^{2} + {(y_{2} - y_{1})}_{2}^{2}}$

Distance between the points A $(0, 0)$ and B $(- 8, 8) = \sqrt{{(- 8 - 0)}^{2} + {(8 - 0)}^{2}} = \sqrt{64 + 64} = \sqrt{128}$

Distance between the points B $(- 8, 8)$ and C $(x, 8) = \sqrt{{(x + 8)}^{2} + {(8 - 8)}^{2}} = \sqrt{{(x + 8)}^{2}}$

Distance between the points A $(0, 0)$ and C $(x, 8) = \sqrt{{(x - 0)}^{2} + {(8 - 0)}^{2}} = \sqrt{x^{2} + 64}$
If the triangle has right angle at A, then
${A B}^{2} + {A C}^{2} = {B C}^{2}$
$=> 128 + x^{2} + 64 = {(x + 8)}^{2}$
$=> 128 + x^{2} + 64 = x^{2} + 64 + 16 x$
$=> 16 x = 128$
$x = 8$

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Let ABC be a right angled triangle whose vertices are A(0,0),B(−8,8), and C(x,8) respectively, then the possible value of x is

Let ABC be a right angled triangle whose vertices are $A (0, 0), B (- 8, 8)$ , and $C (x, 8)$ respectively, then the possible value of x is