The equation of a straight line,perpendicular | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) The given equation of the line is $3x - 4y = 6$. Any line perpendicular to this line is of the form $4x + 3y = k$. Writing this in intercept form,

Vedclass · Accepted Answer

$4x + 3y = 12$

Vedclass · Answer

(B) The given equation of the line is $3x - 4y = 6$. Any line perpendicular to this line is of the form $4x + 3y = k$. Writing this in intercept form,we get $\frac{x}{k/4} + \frac{y}{k/3} = 1$. The intercepts on the coordinate axes are $OA = |\frac{k}{4}|$ and $OB = |\frac{k}{3}|$. The area of the triangle formed with the coordinate axes is given by $\frac{1}{2} 	imes |OA| 	imes |OB| = 6$. $\frac{1}{2} 	imes |\frac{k}{4}| 	imes |\frac{k}{3}| = 6$ $\frac{k^2}{24} = 6$ $k^2 = 144$ $k = \pm 12$. Thus,the required equations are $4x + 3y = 12$ or $4x + 3y = -12$. Comparing with the given options,$4x + 3y = 12$ is the correct choice.

The equation of a straight line,perpendicular to $3x - 4y = 6$ and forming a triangle of area $6$ sq. units with the coordinate axes,is

Similar Questions

Chrysophytes,Euglenoids,Dinoflagellates,and Slime moulds are included in the kingdom:

Find the equivalent capacitance across $A$ and $B$ in $\mu F$.

The sum of the solutions of the equation $|\sqrt{x}-2|+\sqrt{x}(\sqrt{x}-4)+2=0$ for $x>0$ is equal to

The point of intersection of tangents at the ends of the latus rectum of the parabola $y^2 = 4x$ is equal to

When the momentum of a proton is changed by an amount $P_0$,the corresponding change in the de-Broglie wavelength is found to be $0.25\%$. Then,the original momentum of the proton was

Vedclass Test Series

Exam Paper Generator

Online Exam Module