Find the equation of the line passing through the point $(2, 3)$ and perpendicular to the line $3 x + 2 y - 1 = 0$

Open in App

Solution

The line $3 x + 2 y = 1$ is of the form $a x + b y = c$

$\Rightarrow$ slope $= \frac{- a}{b} = \frac{- 3}{2}$

$∴$ slope of the line perpendicular to the line passing through $(2, 3)$ is $= \frac{2}{3}$

$∴$ equation of line passing through $(2, 3)$ is

$\Rightarrow (y - 3) = \frac{2}{3} (x - 2)$

$\Rightarrow 3 y - 9 = 2 x - 4$

$\Rightarrow 2 x - 3 y + 5 = 0$

Suggest Corrections

Similar questions

Q. Find the equation of a line passing through

(2, 3)

and perpendicular to the lines

3 x + 2 y - 1 = 0.

Q. Find the combined equation of lines passing through the point

P (1, 2)

and perpendicular to both the lines

3 x + 2 y - 5 = 0

and

2 x - 5 y + 1 = 0

Find the equation of the straight line passing through the point of intersection of the lines $5 x - 6 y - 1 = 0$ and $3 x + 2 y + 5 = 0$ and perpendicular to the line $3 x - 5 y + 1 = 0.$

Q. The equation of the line perpendicular to the line

2 x + 3 y + 5 = 0

and passing through

(1, 1)

, is

Q. The equation to the straight line passing through the point of intersection of the lines

5 x - 6 y - 1 = 0

and

3 x + 2 y + 5 = 0

and perpendicular to the line

3 x - 5 y + 11 = 0

is?

Join BYJU'S Learning Program

Find the equation of the line passing through the point (2,3) and perpendicular to the line 3x+2y−1=0

The line 3x+2y=1 is of the form ax+by=c ⇒ slope =−ab=−32∴ slope of the line perpendicular to the line passing through (2,3) is =23∴ equation of line passing through (2,3) is⇒(y−3)=23(x−2)⇒3y−9=2x−4⇒2x−3y+5=0

Find the equation of the line passing through the point $(2, 3)$ and perpendicular to the line $3 x + 2 y - 1 = 0$