Question

The in-centre of a triangle whose vertices are $( - 36,7),(20,7)$ and $(0, - 8)$ is ?
A) $(0, - 1)$
B) $( - 1,0)$
C) $(\dfrac{1}{2},1)$
D) None of these

Answer 1

Hint: We can find the length of the sides of the triangle using the vertices given. Distance formulas can be used for this. We can find the coordinates of the in-centre if the vertices and sides of a triangle are known.

Formula used:
Distance formula:
Distance between two points $({x_1},{y_1})$ and $({x_2},{y_2})$ is given by $\sqrt {{{({x_1} - {x_2})}^2} + {{({y_1} - {y_2})}^2}} $.
Suppose the vertices of a triangle are $A({x_1},{y_1}),B({x_2},{y_2})$ and $C({x_3},{y_3})$ and sides opposite to these vertices are $a,b$ and $c$ respectively, then the coordinates of the in-centre is given by
$x = \dfrac{{a{x_1} + b{x_2} + c{x_3}}}{{a + b + c}},y = \dfrac{{a{y_1} + b{y_2} + c{y_3}}}{{a + b + c}}$

Complete step-by-step answer:
Given a triangle with vertices $( - 36,7),(20,7)$ and $(0, - 8)$.

Let it be $A( - 36,7),B(20,7)$ and $C(0, - 8)$ and the sides opposite to these vertices be $a,b$ and $c$.
Distance between two points $({x_1},{y_1})$ and $({x_2},{y_2})$ is given by $\sqrt {{{({x_1} - {x_2})}^2} + {{({y_1} - {y_2})}^2}} $.
So by distance formula we have,
$a = BC = \sqrt {{{(20 - 0)}^2} + {{(7 - - 8)}^2}} = \sqrt {{{20}^2} + {{15}^2}} = \sqrt {400 + 225} = \sqrt {625} = 25$
$b = AC = \sqrt {{{( - 36 - 0)}^2} + {{(7 - - 8)}^2}} = \sqrt {{{( - 36)}^2} + {{15}^2}} = \sqrt {1296 + 225} = \sqrt {1521} = 39$
$c = AB = \sqrt {{{( - 36 - 20)}^2} + {{(7 - 7)}^2}} = \sqrt {{{( - 56)}^2} + {0^2}} = \sqrt {{{56}^2} + 0} = 56$
Now we have to find the in-centre of $\vartriangle ABC$.
Suppose the vertices of a triangle are $A({x_1},{y_1}),B({x_2},{y_2})$ and $C({x_3},{y_3})$ and sides opposite to these vertices are $a,b$ and $c$ respectively, then the coordinates of the in-centre is given by
$x = \dfrac{{a{x_1} + b{x_2} + c{x_3}}}{{a + b + c}},y = \dfrac{{a{y_1} + b{y_2} + c{y_3}}}{{a + b + c}}$
Here the triangle has vertices $( - 36,7),(20,7)$ and $(0, - 8)$.
Also sides are given by $a = 25,b = 39,c = 56$
So substituting we get
$x = \dfrac{{25 \times - 36 + 39 \times 20 + 56 \times 0}}{{25 + 39 + 56}},y = \dfrac{{25 \times 7 + 39 \times 7 + 56 \times - 8}}{{25 + 39 + 56}}$
Simplifying we get,
$x = \dfrac{{25 \times - 36 + 39 \times 20 + 56 \times 0}}{{25 + 39 + 56}},y = \dfrac{{25 \times 7 + 39 \times 7 + 56 \times - 8}}{{25 + 39 + 56}}$
$ \Rightarrow x = \dfrac{{ - 900 + 780 + 0}}{{120}},y = \dfrac{{175 + 273 - 448}}{{120}}$
Solving we get,
$ \Rightarrow x = \dfrac{{ - 120}}{{120}},y = \dfrac{0}{{120}}$
$ \Rightarrow x = - 1,y = 0$
This gives the in-centre is $(x,y) = ( - 1,0)$.
Therefore the answer is option B.

Note: The vertices are given directly in the question. But we need the length of the sides too for finding the in-centre. We had calculated that one using the distance formula. Be careful when substituting the values correctly in the formula for finding the in-centre.