Given:
![\[y\text{ }=\text{ }{{m}_{1}}x\text{ }+\text{ }{{c}_{1}}~\ldots \text{ }\left( 1 \right)\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-5beb3ff438304a3e78ae074565d06eb6_l3.png "Rendered by QuickLaTeX.com")
![\[y\text{ }=\text{ }{{m}_{2}}x\text{ }+\text{ }{{c}_{2}}~\ldots \text{ }\left( 2 \right)\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-f7eda219c50346221a3c26fce6e7c658_l3.png "Rendered by QuickLaTeX.com")
and ,
![\[x\text{ }=\text{ }0\text{ }\ldots \text{ }\left( 3 \right)\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-62d31a14008ae819ab13e8ab609b429b_l3.png "Rendered by QuickLaTeX.com")
In triangle ABC,
let equations (1), (2) and (3) represent the sides AB, BC and CA, respectively.
Solving (1) and (2), we have
Solving (1) and (3):
![\[x\text{ }=\text{ }0,\text{ }y\text{ }=\text{ }{{c}_{1}}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-f74043177b678e4b119301cbea2388cf_l3.png "Rendered by QuickLaTeX.com")
Thus, AB and CA intersect at A 0,c1.
Similarly, solving (2) and (3):
![\[x\text{ }=\text{ }0,\text{ }y\text{ }=\text{ }{{c}_{2}}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-c5157bf0a0dbe4c48994cf22ec5aedc8_l3.png "Rendered by QuickLaTeX.com")
Thus, BC and CA intersect at C 0,c2.
