Putting the value of y1 = 4 – x1 in equation 3, we get
![\[\begin{array}{*{35}{l}} 4{{x}_{1}}~+\text{ }3\left( 4\text{ }-\text{ }{{x}_{1}} \right)\text{ }=\text{ }5 \\ \Rightarrow ~4{{x}_{1}}~+\text{ }12\text{ }-\text{ }3{{x}_{1}}~=\text{ }5 \\ \Rightarrow ~{{x}_{1}}~=\text{ }5\text{ }-\text{ }12 \\ \Rightarrow ~{{x}_{1}}~=\text{ }-\text{ }7 \\ \end{array}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-5d1345c6ecf25868bd3838d21e2bdd9c_l3.png "Rendered by QuickLaTeX.com")
Putting the value of x1 in equation 4, we get
![\[\begin{array}{*{35}{l}} {{y}_{1}}~=\text{ }4\text{ }-\text{ }\left( -7 \right) \\ \Rightarrow ~{{y}_{1}}~=\text{ }4\text{ }+\text{ }7 \\ \Rightarrow ~{{y}_{1}}~=\text{ }11 \\ \end{array}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-eee4351b56cb9d85d44a53b23fad5fde_l3.png "Rendered by QuickLaTeX.com")
Hence, the required points on the given line are (3, 1) and (-7, 11)
Putting the value of y1 = 4 – x1 in equation 3, we get
Putting the value of x1 in equation 4, we get
Hence, the required points on the given line are (3, 1) and (-7, 11)