The equation of the straight line passing through the points of intersection of 2x + 3y + 1 = 0 and 3x − 5y − 5 = 0 is
![\[\begin{array}{*{35}{l}} 2x\text{ }+\text{ }3y\text{ }+\text{ }1\text{ }+~\lambda \left( 3x~-~5y~-~5 \right)\text{ }=\text{ }0 \\ \left( 2\text{ }+\text{ }3\lambda \right)x\text{ }+\text{ }\left( 3~-~5\lambda \right)y\text{ }+\text{ }1~-~5\lambda ~=\text{ }0 \\ y\text{ }=\text{ }\text{ }\left[ \left( 2\text{ }+\text{ }3\lambda \right)\text{ }/\text{ }\left( 3\text{ }\text{ }5\lambda \right) \right]\text{ }\text{ }\left[ \left( 1\text{ }\text{ }5\lambda \right)\text{ }/\text{ }\left( 3\text{ }\text{ }5\lambda \right) \right] \\ \end{array}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-4c8bef726e92819d822568f0ffa24b6a_l3.png "Rendered by QuickLaTeX.com")
The required line is equally inclined to the axes. So, the slope of the required line is either 1 or − 1.
So,
![\[\begin{array}{*{35}{l}} \text{ }\left[ \left( 2\text{ }+\text{ }3\lambda \right)\text{ }/\text{ }\left( 3\text{ }-\text{ }5\lambda \right) \right]\text{ }=\text{ }1\text{ }and\text{ }\text{ }\left[ \left( 2\text{ }+\text{ }3\lambda \right)\text{ }/\text{ }\left( 3\text{ }-\text{ }5\lambda \right) \right]\text{ }=\text{ }-1 \\ -2\text{ }\text{ }-3\lambda ~=\text{ }3\text{ }\text{ }-5\lambda ~and\text{ }2\text{ }+\text{ }3\lambda ~=\text{ }3\text{ }-\text{ }5\lambda \\ \end{array}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-c157f19da2055852cc3ba3b879653f08_l3.png "Rendered by QuickLaTeX.com")
λ = 5/2 and 1/8
Now, substitute the values of λ in (2 + 3λ)x + (3 − 5λ)y + 1 − 5λ = 0, we get the equations of the required lines as:
![\[\begin{array}{*{35}{l}} \left( 2\text{ }+\text{ }15/2 \right)x\text{ }+\text{ }\left( 3\text{ }-\text{ }25/2 \right)y\text{ }+\text{ }1\text{ }\text{ }-25/2\text{ }=\text{ }0\text{ }and\text{ }\left( 2\text{ }+\text{ }3/8 \right)x\text{ }+\text{ }\left( 3\text{ }\text{ }-5/8 \right)y\text{ }+\text{ }1\text{ }\text{ }5/8\text{ }=\text{ }0 \\ 19x\text{ }\text{ }-19y\text{ }\text{ }23\text{ }=\text{ }0\text{ }and\text{ }19x\text{ }+\text{ }19y\text{ }+\text{ }3\text{ }=\text{ }0 \\ \end{array}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-d7453230351bb6682e3cb6a8a51955b0_l3.png "Rendered by QuickLaTeX.com")
∴ The required equation is 19x – 19y – 23 = 0 and 19x + 19y + 3 = 0