SATHEE: Straight Line and Pair of Straight Lines 1 Question 53

Straight Line and Pair of Straight Lines 1 Question 53

53. For points $P = (x_{1}, y_{1})$ and $Q = (x_{2}, y_{2})$ of the coordinate plane, a new distance $d (P, Q)$ is defined by $d (P, Q) = | x_{1} - x_{2} | + | y_{1} - y_{2} |$ .

Let $O = (0, 0)$ and $A = (3, 2)$ . Prove that the set of points in the first quadrant which are equidistant (with respect to the new distance) from $O$ and $A$ consists of the union of a line segment of finite length and an infinite ray. Sketch this set in a labelled diagram. $(2000, 10 M)$

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Solution:

NOTE $d : (P, Q) = | x_{1} - x_{2} | + | y_{1} - y_{2} |$ .

It is new method of representing distance between two points $P$ and $Q$ and in future very important in coordinate geometry.

Now, let $P (x, y)$ be any pont in the first quadrant. We have

$d (P, 0)$	$= \| x - 0 \| + \| y - 0 \| = \| x \| + \| y \| = x + y$
	$[∵ x, y > 0]$
$d (P, A)$	$= \| X - 3 \| + \| Y - 2 \|$	[given]
$d (P, 0)$	$= d (P, A)$	[given]
$\Rightarrow x + y$	$= \| x - 3 \| + \| y - 2 \|$	$\dots (i)$
Case I When $0 < x < 3, 0 < y < 2$

Case I When $0 < x < 3, 0 < y < 2$

In this case, Eq. (i) becomes

$x + y = 3 - x + 2 - y$

$\begin{aligned} \Rightarrow & 2 x + 2 y & = 5 \\ or & x + y & = 5 / 2 \end{aligned}$

Case II When $0 < x < 3, y \geq 2$

Now, Eq. (i) becomes

$\begin{aligned} x + y = 3 - x + y - 2 \\ \Rightarrow 2 x = 1 \\ \Rightarrow x = 1 / 2 \end{aligned}$

Case III When $x \geq 3, 0 < y < 2$

Now, Eq. (i) becomes

$\begin{aligned} x + y = x - 3 + 2 - y \\ \Rightarrow 2 y = - 1 or y = - 1 / 2 \end{aligned}$

Hence, no solution.

Case IV When $x \geq 3, y \geq 2$

In this case, case I changes to

$x + y = x - 3 + y - 2 \Rightarrow 0 = - 5$

which is not possible.

Hence, the solution set is

$$ \begin{aligned} & {(x, y) \mid x=12, y \geq 2} \cup{(x, y)} \mid \ & x+y=5 / 2,0<x<3,02} \end{aligned} $$

The graph is given in adjoining figure.

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Straight Line and Pair of Straight Lines 1 Question 53

53. For points $P = (x_{1}, y_{1})$ and $Q = (x_{2}, y_{2})$ of the coordinate plane, a new distance $d (P, Q)$ is defined by $d (P, Q) = | x_{1} - x_{2} | + | y_{1} - y_{2} |$ .

Solution:

Case IV When $x \geq 3, y \geq 2$

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एनसीईआरटी पुस्तकें और समाधान

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अन्य संसाधन

पाठ्यक्रम

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नमूना मॉक टेस्ट

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एनसीईआरटी व्यायाम वीडियो समाधान

Straight Line and Pair of Straight Lines 1 Question 53

53. For points P=(x1,y1) and Q=(x2,y2) of the coordinate plane, a new distance d(P,Q) is defined by d(P,Q)=|x1−x2|+|y1−y2|.

Solution:

Case IV When x≥3,y≥2

इंजीनियरिंग की तैयारी

चिकित्सा तैयारी

एनसीईआरटी पुस्तकें और समाधान

महत्वपूर्ण संसाधन

अन्य संसाधन

पाठ्यक्रम

परीक्षा मंच

नमूना मॉक टेस्ट

सदस्यता

एनसीईआरटी व्यायाम वीडियो समाधान

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53. For points $P = (x_{1}, y_{1})$ and $Q = (x_{2}, y_{2})$ of the coordinate plane, a new distance $d (P, Q)$ is defined by $d (P, Q) = | x_{1} - x_{2} | + | y_{1} - y_{2} |$ .

Case IV When $x \geq 3, y \geq 2$