The family of straight lines 4ax + 3by + c = | vedclass

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

Question

(C) Given the equation of the family of lines: $4ax + 3by + c = 0$ $(1)$.Given the condition: $a + b + c = 0$, which implies $c = -a - b$ $(2)$.Substi

Vedclass · Accepted Answer

$(\frac{1}{4}, \frac{1}{3})$

Vedclass · Answer

(C) Given the equation of the family of lines: $4ax + 3by + c = 0$ $(1)$.Given the condition: $a + b + c = 0$, which implies $c = -a - b$ $(2)$.Substitute $(2)$ into $(1)$:$4ax + 3by - a - b = 0$Rearrange the terms to group $a$ and $b$:$a(4x - 1) + b(3y - 1) = 0$For this equation to hold for all real constants $a$ and $b$, the coefficients of $a$ and $b$ must be zero independently:$4x - 1 = 0 \implies x = \frac{1}{4}$$3y - 1 = 0 \implies y = \frac{1}{3}$Thus, the lines are concurrent at the point $(\frac{1}{4}, \frac{1}{3})$.

Column $I$	Column $II$
$(A)$ $L_1, L_2, L_3$ are concurrent,if	$(p)$ $k=-9$
$(B)$ One of $L_1, L_2, L_3$ is parallel to at least one of the other two,if	$(q)$ $k=-\frac{6}{5}$
$(C)$ $L_1, L_2, L_3$ form a triangle,if	$(r)$ $k=\frac{5}{6}$
$(D)$ $L_1, L_2, L_3$ do not form a triangle,if	$(s)$ $k=5$

The family of straight lines $4ax + 3by + c = 0$ such that $a + b + c = 0$ (where $a, b, c$ are real constants) are concurrent at the point...

Similar Questions

What is the condition for the points $(a, 0)$,$(0, b)$,and $(1, 1)$ to be collinear?

If the lines $x + 2ay + a = 0$,$x + 3by + b = 0$,and $x + 4cy + c = 0$ are concurrent,then $a, b,$ and $c$ are in which of the following?

If the sum of squares of all real values of $\alpha$,for which the lines $2x - y + 3 = 0$,$6x + 3y + 1 = 0$,and $\alpha x + 2y - 2 = 0$ do not form a triangle is $p$,then the greatest integer less than or equal to $p$ is $.........$

The lines $2x + y - 1 = 0$,$ax + 3y - 3 = 0$,and $3x + 2y - 2 = 0$ are concurrent for

Vedclass Test Series

Exam Paper Generator

Online Exam Module