The line 5x + y - 1 = 0 coincides with one of | vedclass

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

Question

(C) The given equation is $5x^2 + xy - kx - 2y + 2 = 0$. Since this represents a pair of lines and one line is $5x + y - 1 = 0$,we can write the equat

Vedclass · Accepted Answer

$11$

Vedclass · Answer

(C) The given equation is $5x^2 + xy - kx - 2y + 2 = 0$. Since this represents a pair of lines and one line is $5x + y - 1 = 0$,we can write the equation as $(5x + y - 1)(ax + by + c) = 0$.Comparing the coefficients of $y^2$,we see the coefficient is $0$,so the second line must be of the form $ax + c = 0$.Thus,$(5x + y - 1)(ax + c) = 5ax^2 + axy + (5c - a)x + cy - c = 0$.Comparing this with $5x^2 + xy - kx - 2y + 2 = 0$:From the $xy$ term,$a = 1$.From the constant term,$-c = 2$,so $c = -2$.From the $y$ term,$c = -2$ (which matches).From the $x$ term,$-k = 5c - a$.Substituting the values: $-k = 5(-2) - 1 = -10 - 1 = -11$.Therefore,$k = 11$.

The line $5x + y - 1 = 0$ coincides with one of the lines given by $5x^2 + xy - kx - 2y + 2 = 0$. Then the value of $k$ is:

Similar Questions

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

If the sum of slopes of the pair of lines given by $4x^{2} + 2hxy - 7y^{2} = 0$ is equal to the product of the slopes,then $h$ is

For what value of $p$ does the equation $y^2 + xy + px^2 - x - 2y = 0$ represent two straight lines?

The value of $k$ so that the equation $2x^2 + 5xy + 3y^2 + 6x + 7y + k = 0$ represents a pair of straight lines is:

If the sum of the slopes of the lines given by ${x^2} - 2cxy - 7{y^2} = 0$ is four times their product,then $c$ has the value

Vedclass Test Series

Exam Paper Generator

Online Exam Module