Math, asked by sanjusaini1195, 6 months ago

Find the equation of the line which cuts off an intercept 2 on the positive directie
x-axis and an intercept 5 on the negative direction of y-axis.​

Answers

Answered by MaheswariS
2

\textbf{Given:}

\textsf{x-intercept of the line is 2}

\textsf{y-intercept of the line is -5}

\textbf{To find:}

\textsf{Equation of the line}

\textbf{Solution:}

\textbf{Intercept form of line:}

\boxed{\mathsf{\dfrac{x}{a}+\dfrac{y}{b}=1}}

\mathsf{Here,\;a=2\;\;and\;\;b=-5}

\textsf{The required line is}

\mathsf{\dfrac{x}{a}+\dfrac{y}{b}=1}

\mathsf{\dfrac{x}{2}+\dfrac{y}{-5}=1}

\mathsf{\dfrac{-5x+2y}{-10}=1}

\mathsf{-5x+2y=-10}

\mathsf{-5x+2y+10=0}

\implies\boxed{\mathsf{5x-2y-10=0}}

Answered by Anonymous
5

\textbf{Given:}

\textsf{x-intercept of the line is 2}

\textsf{y-intercept of the line is -5}

\textbf{To find:}

\textsf{Equation of the line}

\textbf{Solution:}

\textbf{Intercept form of line:}

\boxed{\mathsf{\dfrac{x}{a}+\dfrac{y}{b}=1}}

\mathsf{Here,\;a=2\;\;and\;\;b=-5}

\textsf{The required line is}

\mathsf{\dfrac{x}{a}+\dfrac{y}{b}=1}

\mathsf{\dfrac{x}{2}+\dfrac{y}{-5}=1}

\mathsf{\dfrac{-5x+2y}{-10}=1}

\mathsf{-5x+2y=-10}

\mathsf{-5x+2y+10=0}

\implies\boxed{\mathsf{5x-2y-10=0}}

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