Question

Find the values of non-negative real numbers $h_1,h_2,h_3,k_1,k_2,k_3$ such that the algebraic sum of the perpendiculars drawn from 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.

• A. $h_1=h_2=h_3=k_1=k_2=k_3=0$
• B. $h_1=h_2=h_3=k_1=k_2=k_3=1$
• C. $h_1=h_2=h_3=k_1=k_2=k_3=2$
• D. $h_1=h_2=h_3=k_1=k_2=k_3=4$

Answer 1

Answer: (A)

$h_1=h_2=h_3=k_1=k_2=k_3=0$

Distance of point $(h.k)$ from line $lx+my+n=0$ is

$\frac{lh+mk+n}{\sqrt{l^2+m^2}}$

Let $ax+by+c=0$ be the variable line

Passes through $(2,1)\Rightarrow 2a+b+c=0$

Now,

According to question,

$\frac{2a+k_{1}b+c}{\sqrt{a^2+b^2}}+\frac{3a+k_{2}b+c}{\sqrt{a^2+b^2}}+\frac{7a+k_{3}b+c}{\sqrt{a^2+b^2}}+\frac{h_{1}a+4b+c}{\sqrt{a^2+b^2}}+\frac{h_{2}a+5b+c}{\sqrt{a^2+b^2}}+\frac{h_{3}a-3b+c}{\sqrt{a^2+b^2}}=0$

$\Rightarrow a[2+3+7+h_1+h_2+h_3]+b[k_1+k_2+k_3+4+5-3]+6c=0$

$\Rightarrow a[12+h_1+h_2+h_3]+b[6+k_1+k_2+k_3]+6c=0$

$\Rightarrow a\frac{[2+h_1+h_2+h_3]}{6}+b\frac{[1+k_1+k_2+k_3]}{6}+c=0$

Comparing with $2a+b+c=0$

$h_1+h_2+h_3=0$ & $k_1+k_2+k_3=0$

$\because h_1,h_2,h_3,k_1,k_2$ & $k_3$ are non negative

$\Rightarrow h_1=h_2=h_3=k_1=k_2=k_3=0$