$2 x + 3 y = 7$ and $2 x + 3 y = 12$ are two parallel staright lines, then where does the point $(1, 3)$ lies.

Open in App

Solution

Given: $2 x + 3 y = 7$ $2 x + 3 y = 12$
Let $L_{1} \to 2 x + 3 y - 7 = 0$ $L_{2} \to 2 x + 3 y - 12 = 0$
Let given point $P (1, 3)$
If $L_{1} (P) \times L_{2} (P) < 0$ Point lies between lines
otherwise not between lies
$L_{1} (P) = 2 (1) + 3 (3) - 7 = 4$
$L_{2} (P) = 2 (1) + 3 (3) - 12 = - 1$
$L_{1} (P) \times L_{2} (P) = 4 \times - 1 = - 4 < 0$
So point lies between the lines.

Suggest Corrections

Similar questions

Q. A point which lies between

2 x + 3 y - 7 = 0

and

2 x + 3 y + 12 = 0

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. Find the quadrant, where point of intersection of the lines

2 x - 3 y = - 3

- 12 = - 4 x + y

lies

Q. Show that the lines are parallel
a)

2 x - 3 y + 7 = 0 a n d 12 y = 8 x - 5

Join BYJU'S Learning Program

2x+3y=7 and 2x+3y=12 are two parallel staright lines, then where does the point (1,3) lies.

Given: 2x+3y=7 2x+3y=12Let L1→2x+3y−7=0 L2→2x+3y−12=0Let given point P(1,3)If L1(P)×L2(P)<0 Point lies between linesotherwise not between liesL1(P)=2(1)+3(3)−7=4L2(P)=2(1)+3(3)−12=−1L1(P)×L2(P)=4×−1=−4<0So point lies between the lines.

$2 x + 3 y = 7$ and $2 x + 3 y = 12$ are two parallel staright lines, then where does the point $(1, 3)$ lies.