Question

The symmetric equation of lines $3 x + 2 y + z - 5 = 0$ and $x + y - 2 z - 3 = 0$, is

• A. $\frac { x - 1 } { 5 } = \frac { y - 4 } { 7 } = \frac { z - 0 } { 1 }$
• B. $\frac { x + 1 } { 5 } = \frac { y + 4 } { 7 } = \frac { z - 0 } { 1 }$
• C. $\frac { x + 1 } { - 5 } = \frac { y - 4 } { 7 } = \frac { z - 0 } { 1 }$
• D. $\frac { x - 1 } { - 5 } = \frac { y - 4 } { 7 } = \frac { z - 0 } { 1 }$

Answer 1

Answer: (C)

$\frac { x + 1 } { - 5 } = \frac { y - 4 } { 7 } = \frac { z - 0 } { 1 }$

Answer 2

Let a, b, c be the d.r.'s of required line

$\therefore$ $3 a + 2 b + c = 0$ and $a + b - 2 c = 0$

$\frac { a } { - 4 - 1 } = \frac { b } { 1 + 6 } = \frac { c } { 3 - 2 }$ or $\frac { a } { - 5 } = \frac { b } { 7 } = \frac { c } { 1 }$

In order to find a point on the required line we put $z = 0$ in the two given equation to obtain, $3 x + 2 y = 5$ and $x + y = 3$. Solving these two equations, we obtain $x = - 1 , y = 4$.

$\therefore$ Co-ordinates of point on required line are $( - 1,4,0 )$. Hence required line is$\frac { x + 1 } { - 5 } = \frac { y - 4 } { 7 } = \frac { z - 0 } { 1 }$.