The equations (b - c)x + (c - a)y + (a - b) = | vedclass

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

Question

(D) Two lines $A_1x + B_1y + C_1 = 0$ and $A_2x + B_2y + C_2 = 0$ are identical if $\frac{A_1}{A_2} = \frac{B_1}{B_2} = \frac{C_1}{C_2} = k$.For the g

Vedclass · Accepted Answer

All the above

Vedclass · Answer

(D) Two lines $A_1x + B_1y + C_1 = 0$ and $A_2x + B_2y + C_2 = 0$ are identical if $\frac{A_1}{A_2} = \frac{B_1}{B_2} = \frac{C_1}{C_2} = k$.For the given equations:$\frac{b^3 - c^3}{b - c} = \frac{c^3 - a^3}{c - a} = \frac{a^3 - b^3}{a - b} = k$Using the identity $x^3 - y^3 = (x - y)(x^2 + xy + y^2)$,we get:$b^2 + bc + c^2 = c^2 + ca + a^2 = a^2 + ab + b^2 = k$Case $1$: If $a = b = c$,the equations become $0 = 0$,which represents the same line.Case $2$: If $a, b, c$ are not all equal,then $b^2 + bc + c^2 = c^2 + ca + a^2 \implies b^2 - a^2 + c(b - a) = 0 \implies (b - a)(b + a + c) = 0$.This implies $b = a$ or $a + b + c = 0$.Since the condition must hold for any of the variables being equal,if $a=b$,$b=c$,or $c=a$,the lines are identical. Thus,all the above are correct.

The equations $(b - c)x + (c - a)y + (a - b) = 0$ and $(b^3 - c^3)x + (c^3 - a^3)y + (a^3 - b^3) = 0$ will represent the same line,if

Similar Questions

The value of $k$ such that the lines $2x - 3y + k = 0$,$3x - 4y - 13 = 0$,and $8x - 11y - 33 = 0$ are concurrent,is

If $a$ and $b$ are two arbitrary constants,then the straight line $(a - 2b)x + (a + 3b)y + 3a + 4b = 0$ will pass through

The equation of the straight line joining the point $(a, b)$ to the point of intersection of the lines $\frac{x}{a} + \frac{y}{b} = 1$ and $\frac{x}{b} + \frac{y}{a} = 1$ is

The least positive value of $t$, so that the lines $x=t+\alpha, y+16=0$ and $y=\alpha x$ are concurrent, is

The straight lines $x+3y-9=0$,$4x+5y-1=0$,and $px+qy+10=0$ are concurrent. If the line $5x+6y+10=0$ passes through the point $(a, b)$,find the point.

Vedclass Test Series

Exam Paper Generator

Online Exam Module