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Question: The lines \(x = a y + b\) , <img src="https://cdn.pureessence.tech/canvas_329.png?top_left_x=1645&t...

The lines x=ay+bx = a y + b , and , are perpendicular to each other, if

A

aa′+cc′=1a a ^ { \prime } + c c ^ { \prime } = 1

B

aa′+cc′=−1a a ^ { \prime } + c c ^ { \prime } = - 1

C

ac+a′c′=1a c + a ^ { \prime } c ^ { \prime } = 1

D

ac+a′c′=−1a c + a ^ { \prime } c ^ { \prime } = - 1

Answer

aa′+cc′=−1a a ^ { \prime } + c c ^ { \prime } = - 1

Explanation

Solution

We have, x=ay+bx = a y + b, z=cy+dz = c y + d

x−ba=y\frac { x - b } { a } = y, z−dc=y\frac { z - d } { c } = y

⇒ x−ba=y−01=z−dc\frac { x - b } { a } = \frac { y - 0 } { 1 } = \frac { z - d } { c } ……(i)

and x=a′y+b′x = a ^ { \prime } y + b ^ { \prime }, z=c′y+d′z = c ^ { \prime } y + d ^ { \prime }

x−b′a′=y\frac { x - b ^ { \prime } } { a ^ { \prime } } = y, z−d′c′=y\frac { z - d ^ { \prime } } { c ^ { \prime } } = y

⇒ x−b′a′=y−01=z−d′c′\frac { x - b ^ { \prime } } { a ^ { \prime } } = \frac { y - 0 } { 1 } = \frac { z - d ^ { \prime } } { c ^ { \prime } } ……(ii)

∵ Given, lines (i) and (ii) are perpendicular

∴ a(a′)+1(1)+c(c′)=0a \left( a ^ { \prime } \right) + 1 ( 1 ) + c \left( c ^ { \prime } \right) = 0, aa′+cc′=−1a a ^ { \prime } + c c ^ { \prime } = - 1