A point which lies between $2 x + 3 y - 7 = 0$ and $2 x + 3 y + 12 = 0$ is

(5, 1)

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(- 1, 3)

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(3, - 5)

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(7, - 1)

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Solution

The correct option is A $(5, 1)$

$L_{1} : 2 x + 3 y - 7 = 0$
$L_{2} : 2 x + 3 y + 12 = 0$
Since, Both lines are porallel to each other. so, for a point (x, y) lies b/w $L_{1}$ and $L_{2}$
$L_{1}(-x, y) L_{2}(x, y)<0 $
On checking the option, wegot
$L_{1} (- 1, 3) L_{2} (- 1, 3)$
$= (- 2 + 9 - 7) (- 2 + 9 + 12)$
$= 0$
$L_{1} (5, 1) L_{2} (5, 1) > 0$
$L_{1} (7, - 1) L_{2} (7, - 1) > 0$
But $L_{1} (3, - 5) L_{2} (3, - 5) < 0$
Hence, (3,-5) is the required point
$Since , Both the lines are parallel to each other. (a, b) lies between L_{1}$
$and L_{2}, then$
$\frac{L_{1} (a, b)}{L_{2} (a, b)} > 0$
$on observing the four options:$
$\frac{L_{1} (5, 1)}{L_{2} (5, 1)} > 0$
$Heree, (5, 1) is the required point.$

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Similar questions

2 x + 3 y = 7

and

2 x + 3 y = 12

are two parallel staright lines, then where does the point

(1, 3)

lies.

Show that the point (3, -5) lies between the parallel lines $2 x + 3 y - 7 = 0$ and $2 x + 3 y + 12 = 0$ and find the equation of lines through (3, -5) cutting the above lines at an angle of $45^{\circ}$ .

The equation of the line parallel to the line 2x – 3y = 1 and passing through the middle point of the line segment joining the points (1, 3) and (1, –7), is:

Q. Solve:

\frac{1}{2 (2 x + 3 y)} + \frac{12}{7 (3 x - 2 y)} = \frac{1}{2}

and

\frac{7}{2 x + 3 y} + \frac{4}{3 x - 2 y} = 2

, where

2 x + 3 y \neq 0

and

3 x - 2 y \neq 0

Q. The equation of the parallel plane lying midway between the parallel planes

2 x - 3 y + 6 z - 7 = 0

and

2 x - 3 y + 6 z + 7 = 0

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A point which lies between 2x+3y−7=0 and 2x+3y+12=0 is

A point which lies between $2 x + 3 y - 7 = 0$ and $2 x + 3 y + 12 = 0$ is