Question

The following figure shows an isosceles triangle $OAB$ with side $OA = sie AB = 5 unis and side OB = 6 units$ .

Find the coordinates of the vertices of the $△ OAB$ .

Accepted Answer

Given:

$OA = AB = 5 units, OB = 6 units$

From the above,
$O (0, 0) and B (6, 0)$

Let the coordinates of $A are (x, y)$

Given, $OA = 5 = O B$

By distance formula between two points $(x_{1}, y_{1}), (x_{2}, y_{2})$ is $\sqrt{{(x_{2} - x_{1})}^{2} + {(y_{2} - y_{1})}^{2}}$

$\Rightarrow \sqrt{{(x - 0)}^{2} + {(y - 0)}^{2}} = 5 \Rightarrow \sqrt{x^{2} + y^{2}} = 5 \Rightarrow x^{2} + y^{2} = 25 - - - - (i)$

Given: $A B = O A$

$\Rightarrow \sqrt{{(x - 0)}^{2} + {(y - 0)}^{2}} = \sqrt{{(x - 6)}^{2} + {(y - 0)}^{2}} \Rightarrow \sqrt{x^{2} + y^{2}} = \sqrt{{(x - 6)}^{2} + y^{2}} \Rightarrow x^{2} + y^{2} = {(x - 6)}^{2} + y^{2} \Rightarrow x^{2} + y^{2} = x^{2} - 12 x + 36 + y^{2} \Rightarrow - 12 x + 36 = 0 \Rightarrow 12 x = 36 \Rightarrow x = \frac{36}{12} ∴ x = 3$

From $(i)$ ,
$\Rightarrow 3^{2} + y^{2} = 25 \Rightarrow y^{2} = 25 - 9 \Rightarrow y^{2} = 16 \Rightarrow y = 4$ '

Hence, the coordinates of the vertices are $O (0, 0), A (3, 4) and B (6, 0)$ .

The following figure shows an isosceles triangle $OAB$ with side $OA = sie AB = 5 unis and side OB = 6 units$ .

Find the coordinates of the vertices of the $△ OAB$ .

Important Questions on Chapterwise Revision-(Stage 2)

Learn and Prepare for any exam you want !

The following figure shows an isosceles triangle OAB with side OA=sie AB=5 unis and side OB=6 units. Find the coordinates of the vertices of the △OAB.

Important Questions on Chapterwise Revision-(Stage 2)

Learn and Prepare for any exam you want !

The following figure shows an isosceles triangle $OAB$ with side $OA = sie AB = 5 unis and side OB = 6 units$ .

Find the coordinates of the vertices of the $△ OAB$ .