Equation of the line perpendicular to the line $6 x + 2 y + 7 = 0$ and which dIvides the line segment joining the points $(5, - 3)$ and $(0, 2)$ internally in the ratio $7 : 4$ is

11 x - 33 y - 14 = 0

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

11 x - 33 y + 14 = 0

No worries! We‘ve got your back. Try BYJU‘S free classes today!

33 x - 11 y - 14 = 0

No worries! We‘ve got your back. Try BYJU‘S free classes today!

33 x - 11 y + 14 = 0

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A $11 x - 33 y - 14 = 0$
The point which divides the join of $(5, - 3)$ and $(0, 2)$ in the ratio $7 : 4$ is

$(\frac{7 (0) + 4 (5)}{7 + 4}, \frac{7 (2) + 4 (- 3)}{7 + 4}) = (\frac{20}{11}, \frac{2}{11})$

The equation of the line passing through $(\frac{20}{11}, \frac{2}{11})$ and perpendicular to $6 x + 2 y + 7 = 0$ is

$2 (x - \frac{20}{11}) - 6 (y - \frac{2}{11}) = 0 \Rightarrow (11 x - 20) - 3 (11 y - 2) = 0 \Rightarrow 11 x - 33 y - 14 = 0$

Suggest Corrections

Similar questions

Q. Equation of the line perpendicular to the line

6 x + 2 y + 7 = 0

and which dIvides the line segment joining the points

(5, - 3)

and

(0, 2)

internally in the ratio

7 : 4

Q. A family of lines is given by

(1 + 2 λ) x + (1 - λ) y + λ = 0,

λ

being the parameter. The line belonging to this family at the maximum distance from the point

(1, 4)

If $θ$ is an angle between the lines given by the equation $6 x^{2} + 5 x y - 4 y^{2} + 7 x + 13 y - 3 = 0$ , then equation of the line passing through the point of intersection of these lines and making an angle $θ$ with the positive x - axis is

Q. If the adjacent sides of a parallelogram are represented by

2 x^{2} - 5 x y + 3 y^{2} = 0

and the equation of one diagonal is

x + y - 2 = 0,

then the equation of the other diagonal is

Equation of the straight line passing through the point of intersection of the lines $3 x + 4 y = 7, x - y + 2 = 0$ and having slope $3$ is

Join BYJU'S Learning Program

Equation of the line perpendicular to the line 6x+2y+7=0 and which dIvides the line segment joining the points (5,−3) and (0,2) internally in the ratio 7:4 is

Equation of the line perpendicular to the line $6 x + 2 y + 7 = 0$ and which dIvides the line segment joining the points $(5, - 3)$ and $(0, 2)$ internally in the ratio $7 : 4$ is