The combined equation of the lines passing th | vedclass

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

Question

(A) The slope of the line $x+y+1=0$ is $m = -1$. Let the slope of the required lines be $m'$.Since the angle between the lines is $45^{\circ}$,we use

Vedclass · Accepted Answer

$xy-4x-3y+12=0$

Vedclass · Answer

(A) The slope of the line $x+y+1=0$ is $m = -1$. Let the slope of the required lines be $m'$.Since the angle between the lines is $45^{\circ}$,we use the formula $	an 	heta = |\frac{m_1 - m_2}{1 + m_1 m_2}|$.$	an 45^{\circ} = |\frac{m' - (-1)}{1 + m'(-1)}| \Rightarrow 1 = |\frac{m'+1}{1-m'}|$.This gives two cases: $\frac{m'+1}{1-m'} = 1$ or $\frac{m'+1}{1-m'} = -1$.Case $1$: $m'+1 = 1-m'$ $\Rightarrow 2m' = 0$ $\Rightarrow m' = 0$.The equation of the line is $y-4 = 0(x-3) \Rightarrow y-4 = 0$.Case $2$: $m'+1 = -(1-m')$ $\Rightarrow m'+1 = -1+m'$ $\Rightarrow 1 = -1$,which implies the line is vertical $(m' = \infty)$.The equation of the line is $x-3 = 0$.The combined equation is $(x-3)(y-4) = 0 \Rightarrow xy-4x-3y+12 = 0$.

The combined equation of the lines passing through the point $(3,4)$ and each making an angle $45^{\circ}$ with the line $x+y+1=0$ is

Similar Questions

The joint equation of the pair of lines passing through the origin and forming an equilateral triangle with the line $y=4$ is

If two sides of a triangle are represented by $x^2 - 7xy + 6y^2 = 0$ and the centroid is $(1, 0)$,then the equation of the third side is

The straight lines represented by the equation $9x^{2}-12xy+4y^{2}=0$ are

One of the lines represented by the equation ${x^2} + 6xy = 0$ is

The value of $\lambda$ with $|\lambda| < 16$ such that $2 x^2-10 x y+12 y^2+5 x+\lambda y-3=0$ represents a pair of straight lines,is

Vedclass Test Series

Exam Paper Generator

Online Exam Module