Match the following:

Column C1                                                            Column C2

(a) The coordinates of the points P and Q on the                                (i) $(3, 1), (– 7, 11)$

line $x + 5 y = 13$ , which are at a distance of 2 units

from the line $12 x – 5 y + 26 = 0$ are

(b) The coordinates of the point on the                                                 (ii) $(3, 1), (– 7, 11)$

line $x + y = 4$ , which is at a unit distance

from the line $4 x + 3 y – 10 = 0$ .

(c) The coordinates of the point on the line                                         (iii) $(\frac{12}{5}, \frac{16}{5}), (1, – 3)$

joining A(–2, 5) and B(3, 1) such that

$A P = P Q = Q B$ are

3 Views | Posted 3 months ago

Asked by Shiksha User

1 Answer

Answered by

Vishal Baghel | Contributor-Level 10

3 months ago

This is a True or False Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} (a) L e t P (x_{1}, y_{1}) b e a n y o n t h e g i v e n l i n e \\ x + 5 y = 13 ∴ x_{1} + 5 y_{1} = 13 \\ o f l i n e 12 x - 5 y + 26 = 0 f r o m t h e (x_{1}, y_{1}) \\ 2 = | \frac{12 x_{1} - 5 y_{1} + 26}{\sqrt[]{{(12)}^{2} + {(- 5)}^{2}}} | \Rightarrow 2 = | \frac{12 x_{1} - (13 - x_{1}) + 26}{13} | \\ \Rightarrow 2 = | \frac{12 x_{1} - 13 + x_{1} + 26}{13} | \Rightarrow 2 = | \frac{13 x_{1} + 13}{13} | \\ \Rightarrow 2 = \pm (x_{1} + 1) \\ \Rightarrow 2 = x_{1} + 1 \Rightarrow x_{1} = 1 (T a k i n g (+) s i g n) \\ a n d 2 = - x_{1} - 1 \Rightarrow x_{1} = - 3 (T a k i n g (-) s i g n) \\ P u t t i n g t h e v a l u e o f x_{1} i n e q n . x_{1} + 5 y_{1} = 13 \\ W e g e t y_{1} = \frac{12}{5} a n d \frac{16}{5} \\ S o, t h e r e q u i r e d a r e (1, \frac{12}{5}) a n d (- 3, \frac{16}{5}) . \\ H e n c e, (a) \leftrightarrow (i i i) \\ (b) L e t P (x_{1}, y_{1}) b e a n y o n t h e g i v e n l i n e \\ x + y = 4 ∴ x_{1} + y_{1} = 4 \dots (i) \\ o f l i n e 4 x + 3 y - 10 = 0 f r o m t h e (x_{1}, y_{1}) \\ 1 = | \frac{4 x_{1} + 3 y_{1} - 10}{\sqrt[]{{(4)}^{2} + {(3)}^{2}}} | \Rightarrow 1 = | \frac{4 x_{1} + 3 (4 - x_{1}) - 10}{5} | \\ \Rightarrow 1 = | \frac{4 x_{1} + 12 - 3 x_{1} - 10}{5} | \Rightarrow 1 = | \frac{x_{1} + 2}{5} | \\ \Rightarrow 1 = \pm (\frac{x_{1} + 2}{5}) \\ \Rightarrow \frac{x_{1} + 2}{5} = 1 (T a k i n g (+) s i g n) \\ \Rightarrow x_{1} + 2 = 5 \Rightarrow x_{1} = 3 \\ a n d \frac{x_{1} + 2}{5} = - 1 (T a k i n g (-) s i g n) \\ \Rightarrow x_{1} + 2 = - 5 \Rightarrow x_{1} = - 7 \\ P u t t i n g t h e v a l u e o f x_{1} i n e q n . (i) w e g e t \\ A t x_{1} = 3, y_{1} = 1 \\ A t x_{1} = - 7, y_{1} = 11 \\ S o, t h e r e q u i r e d a r e (3, 1) a n d (- 7, 11) . \\ H e n c e, (b) \leftrightarrow (i) . \end{array}$

$\begin{array}{l} (c) G i v e n t h a t A P = P Q = Q B \\ E q u a t i o n o f l i n e j o i n i n g A (- 2, 5) a n d B (3, 1) i s \\ y - 5 = \frac{1 - 5}{3 + 2} (x + 2) \\ \Rightarrow y - 5 = \frac{- 4}{5} (x + 2) \\ \Rightarrow 5 y - 25 = - 4 x - 8 \\ \Rightarrow 4 x + 5 y - 17 = 0 \\ L e t P (x_{1}, y_{1}) a n d Q (x_{2}, y_{2}) b e a n y t w o points o n t h e l i n e A B \\ P (x_{1}, y_{1}) d i v i d e s t h e l i n e A B i n t h e r a t i o 1 : 2 \\ ∴ x_{1} = \frac{1.3 + 2 (- 2)}{1 + 2} = \frac{3 - 4}{3} = \frac{- 1}{3} \\ y_{1} = \frac{1.1 + 2.5}{1 + 2} = \frac{1 + 10}{3} = \frac{11}{3} \\ S o, t h e c o o r d i n a t e s o f P (x_{1}, y_{1}) = (\frac{- 1}{3}, \frac{11}{3}) . \\ N o w point Q (x_{2}, y_{2}) i s t h e m i d -point o f P B \\ ∴ x_{2} = \frac{3 - \frac{1}{3}}{2} = \frac{4}{3} \\ y_{2} = \frac{1 + \frac{11}{3}}{2} = \frac{7}{3} \\ H e n c e, t h e c o o r d i n a t e s o f Q (x_{2}, y_{2}) = (\frac{4}{3}, \frac{7}{3}) . \\ H e n c e, (c) \leftrightarrow (i i) . \end{array}$

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