Question

If $p_1,p_2,p_3$ are altitudes of a triangle $ABC$ from the vertices $A,B,C$ and $△$ the area of the triangle,then $p_1^{-1}+p_2^{-1}-p_3^{-1}$ is equal to

• A. $\frac{s-a}{△}$
• B. $\frac{s-b}{△}$
• C. $\frac{s-c}{△}$
• D. $\frac{s}{△}$

Answer 1

The correct option is C $\frac{s-c}{△}$

$\because △=\frac{1}{2}.a.p_1$
Rearranging, we get
$\Rightarrow p_1=\frac{2△}{a}$
$\Rightarrow \frac{1}{p_1}=\frac{a}{2△}$ .............$\left(1\right)$
$△=\frac{1}{2}.b.p_2$
Rearranging, we get
$\Rightarrow p_2=\frac{2△}{b}$
$\Rightarrow \frac{1}{p_2}=\frac{b}{2△}$ .............$\left(2\right)$
and $△=\frac{1}{2}.c.p_3$
Rearranging, we get
$\Rightarrow p_3=\frac{2△}{c}$
$\Rightarrow \frac{1}{p_3}=\frac{c}{2△}$ ..............$\left(3\right)$
Adding $\left(1\right),\left(2\right),$ and $\left(3\right)$ we get
$\frac{1}{p_1}+\frac{1}{p_2}-\frac{1}{p_3}=\frac{a}{2△}+\frac{b}{2△}-\frac{c}{2△}$
$=\frac{a+b-c}{2△}$
$=\frac{2s-c-c}{2△}$
$=\frac{2s-2c}{2△}$
$=\frac{s-c}{△}$