p_1, p_2, p_3 are altitudes of a triangle ABC | vedclass

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(C) We know that the area of the triangle $\Delta = \frac{1}{2} a p_1 = \frac{1}{2} b p_2 = \frac{1}{2} c p_3$. From this,we get $\frac{1}{p_1} = \fra

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$\frac{s-c}{\Delta}$

Vedclass · Answer

(C) We know that the area of the triangle $\Delta = \frac{1}{2} a p_1 = \frac{1}{2} b p_2 = \frac{1}{2} c p_3$. From this,we get $\frac{1}{p_1} = \frac{a}{2\Delta}$,$\frac{1}{p_2} = \frac{b}{2\Delta}$,and $\frac{1}{p_3} = \frac{c}{2\Delta}$. Substituting these into the expression: $\frac{1}{p_1} + \frac{1}{p_2} - \frac{1}{p_3} = \frac{a}{2\Delta} + \frac{b}{2\Delta} - \frac{c}{2\Delta} = \frac{a+b-c}{2\Delta}$. Since $2s = a+b+c$,we have $a+b = 2s-c$. Therefore,$\frac{a+b-c}{2\Delta} = \frac{(2s-c)-c}{2\Delta} = \frac{2s-2c}{2\Delta} = \frac{s-c}{\Delta}$.

$p_1, p_2, p_3$ are altitudes of a triangle $ABC$ drawn from the vertices $A, B, C$ respectively. If $\Delta$ is the area of the triangle and $2s$ is the sum of the sides,then $\frac{1}{p_1} + \frac{1}{p_2} - \frac{1}{p_3} =$

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