First slide

Equation of line in Straight lines

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.

Moderate

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

$h_{1} = h_{2} = h_{3} = k_{1} = k_{2} = k_{3} = 4$

Solution

Let the equation of variable line be $a x + b y + c = 0$ , it is given that $\sum_{i = 1}^{6} \frac{a x_{i} + b y_{i} + c}{\sqrt{a^{2} + b^{2}}} = 0$ $\Rightarrow a (\frac{\sum x_{i}}{6}) + b (\frac{\sum y_{i}}{6}) + c = 0$
So, the fixed point must be $\frac{\sum x_{i}}{6}$ , $\frac{\sum y_{i}}{6}$ . But fixed point is (2,1) so $(2 + 3 + 7 + h_{1} + h_{2} + h_{3}) / 6 = 2$
$\Rightarrow h_{1} + h_{2} + h_{3} = 0$
$\Rightarrow h_{1} = 0, h_{2} = 0, h_{3} = 0$
(as $h_{1}, h_{2}, h_{3}$ are non-negative) similarly, we get
$\frac{k_{1} + k_{2} + k_{3} + 4 + 5 + - 3}{6} = 1$
$\Rightarrow k_{1} = k_{2} = k_{3} = 0$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App