Question

For non-negative real numbers $h_1,h_2,h_3,k_1,k_2,k_3,$ if the algebraic sum of the perpendiculars drawn from the points $(2,k_1),$ $(3,k_2),$ $(7,k_3),$ $(h_1,4),$ $(h_2,5),$ $(h_3,-3)$ on a variable line passing through $(2,1)$ is zero, then

• A. $h_1=1,h_2=1,h_3=2$
• B. $k_1=k_2=k_3=0$
• C. $h_1=h_2=h_3=0$
• D. $k_1=1,k_2=2,k_3=3$

Answer 1

Answer: (B), (C)

$h_1=h_2=h_3=0$

Let the equation of the variable line be $ax+by+c=0$. It is given that
$\sum _{i=1}^6\frac{a{x}_i+b{y}_i+c}{\sqrt{a^2+b^2}}=0(\because \text{algebraic sum of perpendiculars from given six points is}0)\Rightarrow \sum _{i=1}^6(a{x}_i+b{y}_j+c)=0$
$\Rightarrow \sum _{i=1}^{6}a{x}_i+\sum _{i=1}^{6}b{y}_i+\sum _{i=1}^{6}c=0\Rightarrow \sum _{i=1}^{6}a{x}_i+\sum _{i=1}^{6}b{y}_i+6c=0$
$\Rightarrow a\left(\frac{\sum x_i}{6}\right)+b\left(\frac{\sum y_i}{6}\right)+c=0$

Hence, the line $ax+by+c=0$ always passes through point $(\frac{\sum x_i}{6},\frac{\sum y_i}{6})$
But it is given line passes through point $(2,1)$
Hence, $\frac{2+3+7+h_1+h_2+h_3}{6}=2$
$\Rightarrow h_1+h_2+h_3=0$
$\Rightarrow h_1=0,h_2=0,h_3=0$
(as $h_1,h_2,h_3$ are non-negative).

Similarly, $\frac{k_1+k_2+k_3+4+5-3}{6}=1$
$\Rightarrow k_1=k_2=k_3=0$