For all values of a and b,the line (a+2b)x + | vedclass

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

Question

(C) The given equation is $(a+2b)x + (a-b)y + (a+5b) = 0$. Rearranging the terms by grouping $a$ and $b$: $a(x + y + 1) + b(2x - y + 5) = 0$. For this

Vedclass · Accepted Answer

$(-2, 1)$

Vedclass · Answer

(C) The given equation is $(a+2b)x + (a-b)y + (a+5b) = 0$. Rearranging the terms by grouping $a$ and $b$: $a(x + y + 1) + b(2x - y + 5) = 0$. For this equation to hold for all values of $a$ and $b$,the coefficients of $a$ and $b$ must be zero independently: $x + y + 1 = 0$  --- $(1)$ $2x - y + 5 = 0$ --- $(2)$ Adding equations $(1)$ and $(2)$: $(x + y + 1) + (2x - y + 5) = 0$ $3x + 6 = 0$ $3x = -6$ $x = -2$. Substituting $x = -2$ into equation $(1)$: $-2 + y + 1 = 0$ $y - 1 = 0$ $y = 1$. Thus,the line passes through the fixed point $(-2, 1)$.

For all values of $a$ and $b$,the line $(a+2b)x + (a-b)y + (a+5b) = 0$ passes through a fixed point. Find that point.

Similar Questions

$A$ student has to answer $10$ out of $13$ questions in an examination,choosing at least $5$ questions from the first $6$ questions. The number of choices available to the student is:

$3.011 \times 10^{22}$ atoms of an element weigh $1.15 \ g$. The atomic mass of the element is

$ABCD$ is a square with side $16$ units and $A$ is the origin. If the equation of the circle circumscribing the square $ABCD$ is $x^2+y^2=4k(x+y)$,then $k=$

The reading of the ammeter in the circuit shown will be: (in $A$)

How many cyclic structures are possible for $C_4H_6$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module