The lines 15x - 18y + 1 = 0,12x + 10y - 3 = 0 | vedclass

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

Question

(C) Let the lines be $L_1: 15x - 18y + 1 = 0$,$L_2: 12x + 10y - 3 = 0$,and $L_3: 6x + 66y - 11 = 0$.Three lines are concurrent if there exist constant

Vedclass · Accepted Answer

Concurrent

Vedclass · Answer

(C) Let the lines be $L_1: 15x - 18y + 1 = 0$,$L_2: 12x + 10y - 3 = 0$,and $L_3: 6x + 66y - 11 = 0$.Three lines are concurrent if there exist constants $a$ and $b$ such that $L_3 = aL_1 + bL_2$.Consider the linear combination $3(12x + 10y - 3) - 2(15x - 18y + 1) = (36x - 30x) + (30y + 36y) + (-9 - 2) = 6x + 66y - 11$.This matches $L_3 = 0$.Since $L_3$ can be expressed as a linear combination of $L_1$ and $L_2$,the lines are concurrent.

The lines $15x - 18y + 1 = 0$,$12x + 10y - 3 = 0$ and $6x + 66y - 11 = 0$ are

Similar Questions

What is the nature of the points $(-a, -b)$,$(0, 0)$,$(a, b)$,and $(a^2, ab)$?

If the three lines $2x + y = 4$,$3x + 2y = 3$,and $ax + 3y = 2$ are concurrent,find the value of $a$.

The system of equations $2x + y - 5 = 0$,$x - 2y + 1 = 0$,and $2x - 14y - a = 0$ is consistent. Then,$a$ is equal to

If the lines $3x + y - 2 = 0$,$px + 2y - 3 = 0$,and $2x - y - 3 = 0$ are concurrent,then $p =$

The equation of the line passing through the point of intersection of the lines $3x - 2y - 1 = 0$ and $x - 4y + 3 = 0$ and the point $(\pi, 0)$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module