Question

The point of concurrence of the lines X/3+y/4=1, X/4+y/3=1, X=y is​

Answer 1

$\textbf{Given:}$

$\textsf{Lines are}$

$\mathsf{\dfrac{x}{3}+\dfrac{y}{4}=1,\;\dfrac{x}{4}+\dfrac{y}{3}=1,\;x=y}$

$\textbf{To find:}$

$\textsf{The point of concurrece of the given lines}$

$\textbf{Solution:}$

$\textsf{The point of concurrence of the given lines is obtained by}$

$\textsf{solving the equations}$

$\mathsf{\dfrac{x}{3}+\dfrac{y}{4}=1}$.......(1)

$\mathsf{\dfrac{x}{4}+\dfrac{y}{3}=1}$......(2)

$\mathsf{x=y}$......(3)

$\textsf{Using (3) in (1)}$

$\mathsf{\dfrac{y}{3}+\dfrac{y}{4}=1}$

$\mathsf{\dfrac{4y+3y}{12}=1}$

$\mathsf{\dfrac{7y}{12}=1}$

$\implies\mathsf{y=\dfrac{12}{7}}$

$\mathsf{(3)\implies\;x=\dfrac{12}{7}}$

$\textsf{Similarly, by solving (2) and (3) we get}$

$\mathsf{x=\dfrac{12}{7}\;\;and\;\;y=\dfrac{12}{7}}$

$\therefore\mathsf{The\;point\;of\;concurrence \;is\;\left(\dfrac{12}{7},\dfrac{12}{7}\right)}$

$\textbf{Find more:}$

Find value of k if the lines 3x-4y-13=0, 8x-11y-33=0, 2x-3y+k=0 are concurrent

https://brainly.in/question/2837292

Show that the lines 2x+y-3=0, 3x+2y-2=0, 2x-3y-23=0 are concurrent.Also find the point of concurrency.​

https://brainly.in/question/11502159

Answer 2

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