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In the given figure, point D divides the side BC of $\triangle ABC$ in the ratio $1:2$. Find length AD.
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Sol. Coordinates of point D = $(\frac{1\times 4+2\times (-2)}{1+2}, \frac{1\times 2+2\times 1}{1+2})$ i.e. $(0,\frac{4}{3})$ (1 Mark)
AD = $\sqrt{(1-0)^2 + (5-\frac{4}{3})^2} = \frac{\sqrt{130}}{3}$ units (1 Mark)
AD = $\sqrt{(1-0)^2 + (5-\frac{4}{3})^2} = \frac{\sqrt{130}}{3}$ units (1 Mark)