SOLUTION:
The correct option is option(a)
Explanation:
Given two lines are
![\[\begin{array}{*{35}{l}} 2x\text{ }-\text{ }3y\text{ }+\text{ }5\text{ }=\text{ }0~\ldots \text{ }\left( i \right) \\ ~\text{ }~\text{ }~\text{ }~\text{ }~\text{ }~\text{ }~\text{ }~\text{ }~\text{ }~\text{ }~\text{ }~\text{ }~\text{ }~\text{ }~\text{ }~3x\text{ }+\text{ }4y\text{ }=\text{ }0\text{ }\ldots \text{ }\left( ii \right) \\ \end{array}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-84c90756690063ca3ce10e7e1f6d2de0_l3.png "Rendered by QuickLaTeX.com")
Now, point of intersection of these lines can be find out as
Multiplying equation (i) by 3, we get
![\[6x\text{ }-\text{ }9y\text{ }+\text{ }15\text{ }=\text{ }0\text{ }\ldots \text{ }\left( iii \right)\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-45aaaf4602cad6f19cd390f9f163ef62_l3.png "Rendered by QuickLaTeX.com")
Multiplying equation (ii) by 2, we get
![\[6x\text{ }+\text{ }8y\text{ }=\text{ }0\text{ }\ldots \text{ }\left( iv \right)\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-08cf434239be73d5b72cede5bce03f35_l3.png "Rendered by QuickLaTeX.com")
On subtracting equation (iv) from (iii), we get
![\[\begin{array}{*{35}{l}} 6x\text{ }-\text{ }9y\text{ }+\text{ }15\text{ }-\text{ }6x\text{ }-\text{ }8y\text{ }=\text{ }0 \\ \Rightarrow ~\text{ }-17y\text{ }+\text{ }15\text{ }=\text{ }0 \\ \Rightarrow ~\text{ }-17y\text{ }=\text{ }-15 \\ \end{array}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-9092d74e8c4c2704eb6ebb026fb784e7_l3.png "Rendered by QuickLaTeX.com")
