Find equation to $a$ straight line passing through point $(4, - 5)$ & $⊥$ straight line $3 x + 4 y + 5 = 0$

Open in App

Solution

The equation of line perpendicular to the straight line $3 x + 4 y + 5 = 0$ is $4 x - 3 y = c$ .......(1) [ Where $c$ being constant].
Now according to the problem, the line (1) passes through $(4, - 5)$ then,
$16 + 15 = c$
or, $c = 31$ .
So the equation of the line is $4 x - 3 y = 31$ .

Suggest Corrections

Similar questions

Q. Find the equation of the straight line that passes through the point (3, 4) and perpendicular to the line 3x + 2y + 5 = 0

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

5 x - 6 y = 1

and

3 x + 2 y + 5 = 0

and is perpendicular to the straight line

3 x - 5 y + 11 = 0

Equation of a straight line passing through the point $(4, 5)$ and equally inclined to the lines $3 x = 4 y + 7$ and $5 y = 12 x + 6$ is

Find the equation of a straight line passing through the point of intersection of x+2y+3=0 and 3x+4y+7=0 and perpenicular to the straight line x-y+0=0

Find the equation of a line passing through the point $(2, 3)$ and parallel to the line $3 x - 4 y + 5 = 0.$

Join BYJU'S Learning Program

Find equation to a straight line passing through point (4,−5) & ⊥ straight line 3x+4y+5=0

The equation of line perpendicular to the straight line 3x+4y+5=0 is 4x−3y=c.......(1) [ Where c being constant].Now according to the problem, the line (1) passes through (4,−5) then,16+15=cor, c=31.So the equation of the line is 4x−3y=31.

Find equation to $a$ straight line passing through point $(4, - 5)$ & $⊥$ straight line $3 x + 4 y + 5 = 0$

The equation of line perpendicular to the straight line $3 x + 4 y + 5 = 0$ is $4 x - 3 y = c$ .......(1) [ Where $c$ being constant].
Now according to the problem, the line (1) passes through $(4, - 5)$ then,
$16 + 15 = c$
or, $c = 31$ .
So the equation of the line is $4 x - 3 y = 31$ .