The pair of straight lines that passes throug | vedclass

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

Question

(B) The given equation of the pair of straight lines is $3x^2 - 8xy + 5y^2 = 0$. Factoring this,we get $3x^2 - 3xy - 5xy + 5y^2 = 0$,which simplifies

Vedclass · Accepted Answer

$(5x + 3y - 11)(x + y - 3) = 0$

Vedclass · Answer

(B) The given equation of the pair of straight lines is $3x^2 - 8xy + 5y^2 = 0$. Factoring this,we get $3x^2 - 3xy - 5xy + 5y^2 = 0$,which simplifies to $3x(x - y) - 5y(x - y) = 0$,or $(3x - 5y)(x - y) = 0$. Thus,the individual lines are $L_1: 3x - 5y = 0$ and $L_2: x - y = 0$. $A$ line perpendicular to $ax + by + c = 0$ is of the form $bx - ay + k = 0$. For $L_1: 3x - 5y = 0$,the perpendicular line is $5x + 3y + k_1 = 0$. Since it passes through $(1, 2)$,$5(1) + 3(2) + k_1 = 0 \implies 5 + 6 + k_1 = 0 \implies k_1 = -11$. So,$5x + 3y - 11 = 0$. For $L_2: x - y = 0$,the perpendicular line is $x + y + k_2 = 0$. Since it passes through $(1, 2)$,$1 + 2 + k_2 = 0 \implies k_2 = -3$. So,$x + y - 3 = 0$. The combined equation is $(5x + 3y - 11)(x + y - 3) = 0$.

The pair of straight lines that passes through the point $(1, 2)$ and is perpendicular to the pair of straight lines $3x^2 - 8xy + 5y^2 = 0$ is:

Similar Questions

The equation of the pair of straight lines,each of which makes an angle $\alpha$ with the line $y = x$,is

The joint equation of two lines passing through the origin,each making an angle of $30^{\circ}$ with the positive $Y$-axis,is

The combined equation of the pair of lines passing through the origin and making an angle of $\frac{\pi}{4}$ with the line $3x + y - 6 = 0$ is...

Find the length of the intercept cut by the pair of lines $2x^2 + 4xy - 4y^2 - 6x - 8y + 7 = 0$ on the $Y$-axis.

Find the equation of the pair of lines passing through the origin and perpendicular to the pair of lines given by $5x^2 - 7xy - 3y^2 = 0$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module