The equation of the line passing through the | vedclass

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

Question

(C) The equation of the line passing through the point of intersection of the lines $3x - 4y + 1 = 0$ and $5x + y - 1 = 0$ is given by $(3x - 4y + 1)

Vedclass · Accepted Answer

$23x + 23y = 11$

Vedclass · Answer

(C) The equation of the line passing through the point of intersection of the lines $3x - 4y + 1 = 0$ and $5x + y - 1 = 0$ is given by $(3x - 4y + 1) + \lambda(5x + y - 1) = 0$.Rearranging the terms,we get $(3 + 5\lambda)x + (\lambda - 4)y = \lambda - 1$.The intercept form of the line is $\frac{x}{\frac{\lambda - 1}{3 + 5\lambda}} + \frac{y}{\frac{\lambda - 1}{\lambda - 4}} = 1$.Since the intercepts are equal and non-zero,we have $\frac{\lambda - 1}{3 + 5\lambda} = \frac{\lambda - 1}{\lambda - 4}$ with $\lambda 
eq 1$.This implies $\lambda - 4 = 3 + 5\lambda$,which gives $4\lambda = -7$,so $\lambda = -\frac{7}{4}$.Substituting $\lambda = -\frac{7}{4}$ into the equation $(3 + 5\lambda)x + (\lambda - 4)y = \lambda - 1$,we get $(3 - \frac{35}{4})x + (-\frac{7}{4} - 4)y = -\frac{7}{4} - 1$.Simplifying,$-\frac{23}{4}x - \frac{23}{4}y = -\frac{11}{4}$,which results in $23x + 23y = 11$.

The equation of the line passing through the point of intersection of the lines $3x - 4y + 1 = 0$ and $5x + y - 1 = 0$ and making equal non-zero intercepts on the coordinate axes is

Similar Questions

$P$ is a point on either of the two lines $y - \sqrt{3}|x| = 2$ at a distance of $5 \ units$ from their point of intersection. The coordinates of the foot of the perpendicular from $P$ on the bisector of the angle between them are

$A$ pair of straight lines drawn through the origin form an isosceles right-angled triangle with the line $2x + 3y = 6$. Find the equations of the lines and the area of the triangle thus formed.

$A$ straight line passing through $P(3, 1)$ meets the coordinate axes at $A$ and $B$. It is given that the distance of this straight line from the origin $O$ is maximum. The area of triangle $OAB$ is equal to

The equation of the straight line perpendicular to $5x - 2y = 7$ and passing through the point of intersection of the lines $2x + 3y = 1$ and $3x + 4y = 6$ is

The equation of the line joining the point $(3, 5)$ to the point of intersection of the lines $4x + y - 1 = 0$ and $7x - 3y - 35 = 0$ is equidistant from the points $(0, 0)$ and $(8, 34)$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module