The equation of the line which is perpendicular to x -4 y -5-0 at it-s y interceptA- 16 x-4 y-5-0B- 16 x-4 y-5-0C- 16 x-4 y-5-0D- 16 x-4 y-5-0

The correct option is D 16x 4y+5=0y intercept of the given line will be 54 slope of the perpendicular line will be 4 nbsp; So, the required equation is nbs ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Mathematics

Condition for Two Lines to Be Perpendicular

The equation ...

Question

The equation of the line which is perpendicular to $x + 4 y - 5 = 0$ at it's $y$ intercept

16 x + 4 y - 5 = 0

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

16 x - 4 y - 5 = 0

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

16 x + 4 y + 5 = 0

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

16 x - 4 y + 5 = 0

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

Open in App

Solution

The correct option is D $16 x - 4 y + 5 = 0$
$y$ intercept of the given line will be $\frac{5}{4}$
slope of the perpendicular line will be $4$
So, the required equation is
$y = 4 x + \frac{5}{4} \Rightarrow 16 x - 4 y + 5 = 0$

Suggest Corrections

Similar questions

Q. If the two lines

x + y = 6

and

x + 2 y = 4

are the diameters of the circle which passes through

(2, 6)

, then its equation is

Q. Equation(s) of the line perpendicular to

x = y

and at a distance of

1

unit from origin is/are

Q. Equation of straight line midway between the lines

x + 2 y - 1 = 0

and

2 x + 4 y - 5 = 0

Q. The equation of the line which is perpendicular to

x + 4 y - 5 = 0

at it's

y

intercept

Find the condition that the point P(x, y) may lie on the line joining (3, 4) and (-5, -6).

Join BYJU'S Learning Program

The equation of the line which is perpendicular to x+4y−5=0 at it's y intercept

The correct option is D 16x−4y+5=0y intercept of the given line will be 54 slope of the perpendicular line will be 4 So, the required equation is y=4x+54⇒16x−4y+5=0

The equation of the line which is perpendicular to $x + 4 y - 5 = 0$ at it's $y$ intercept

The correct option is D $16 x - 4 y + 5 = 0$
$y$ intercept of the given line will be $\frac{5}{4}$
slope of the perpendicular line will be $4$
So, the required equation is
$y = 4 x + \frac{5}{4} \Rightarrow 16 x - 4 y + 5 = 0$