For the straight lines given by the equation | vedclass

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

Question

(B) The given equation is $(2 + k)x + (1 + k)y = 5 + 7k$. Rearranging the terms to separate $k$: $2x + kx + y + ky = 5 + 7k$ $(2x + y - 5) + k(x + y -

Vedclass · Accepted Answer

Lines pass through the point $(-2, 9)$

Vedclass · Answer

(B) The given equation is $(2 + k)x + (1 + k)y = 5 + 7k$. Rearranging the terms to separate $k$: $2x + kx + y + ky = 5 + 7k$ $(2x + y - 5) + k(x + y - 7) = 0$. This represents a family of lines passing through the intersection of the lines $2x + y - 5 = 0$ and $x + y - 7 = 0$. Solving these two equations: $2x + y = 5$ $x + y = 7$ Subtracting the second from the first: $(2x - x) + (y - y) = 5 - 7$,which gives $x = -2$. Substituting $x = -2$ into $x + y = 7$: $-2 + y = 7$,so $y = 9$. Thus,all lines pass through the point $(-2, 9)$.

For the straight lines given by the equation $(2 + k)x + (1 + k)y = 5 + 7k$,for different values of $k$,which of the following statements is true?

Similar Questions

The lines
$(p - q)x + (q - r)y + (r - p) = 0$
$(q - r)x + (r - p)y + (p - q) = 0$
$(r - p)x + (p - q)y + (q - r) = 0$ are

If the lines $x+3y-5=0$,$5x+2y-12=0$,and $3x-ky-1=0$ do not form a triangle,then a value of $k$ is

If the lines $3x + 4y - 5 = 0$,$2x + 3y - 4 = 0$,and $px + 4y - 6 = 0$ all meet at the same point,then $p$ is equal to:

Find the equation of the line passing through the point of intersection of the lines $x + y - 2 = 0$ and $2x - y + 1 = 0$ and the origin $(0, 0)$.

The lines $x+2ay+a=0$,$x+3by+b=0$,and $x+4cy+c=0$ are concurrent. Then $a, b, c$ are in:

Vedclass Test Series

Exam Paper Generator

Online Exam Module