Given:
![\[y\text{ }=\text{ }\surd 3x\text{ }+\text{ }1\ldots \ldots \text{ }\left( 1 \right)\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-fcda7c801bbce0b22250765ab3ac79e2_l3.png "Rendered by QuickLaTeX.com")
![\[y\text{ }=\text{ }4\text{ }\ldots \ldots .\text{ }\left( 2 \right)\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-fdbce83210003a67f019f07f4043739b_l3.png "Rendered by QuickLaTeX.com")
and
![\[y\text{ }=\text{ }\text{ }\surd 3x\text{ }+\text{ }2\ldots \ldots .\text{ }\left( 3 \right)\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-094eaa7d0591edfd8c3570dc396a0b33_l3.png "Rendered by QuickLaTeX.com")
assume in triangle ABC,
equations (1), (2) and (3) represent the sides AB, BC and CA, respectively.
By solving equations (1) and (2), we have
![\[x\text{ }=~\surd 3,\text{ }y\text{ }=\text{ }4\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-56c7890f40225d82013f985520700a06_l3.png "Rendered by QuickLaTeX.com")
Thus,
AB and BC intersect at B(√3,4)
Now, solving equations (1) and (3), we have
![\[x\text{ }=\text{ }1/2\surd 3,\text{ }y\text{ }=\text{ }3/2\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-3ca57acc4a3009d598e9ab687e556503_l3.png "Rendered by QuickLaTeX.com")
Thus, AB and CA intersect at A (1/2√3, 3/2)
Similarly, solving equations (2) and (3), we get
![\[x\text{ }=\text{ }-2/\surd 3,\text{ }y\text{ }=\text{ }4\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-18b7761ccce3035f14a606414aa5320f_l3.png "Rendered by QuickLaTeX.com")
Thus, BC and AC intersect at C (-2/√3,4)
Now, we have:
Hence proved,
the given lines form an equilateral triangle.