# Co-ordinate Geometry - 2

Some important points in a Triangle:
Centroid: If $\left( {{x_{1,}}{y_1}} \right)$, $\left( {{x_{2,}}{y_2}} \right)$ and $\left( {{x_{3,}}{y_3}} \right)$ are the vertices of a triangle, then the coordinates of its centroid are $\left( {\displaystyle\frac{{{x_1} + {x_2} + {x_3}}}{3},\frac{{{y_1} + {y_2} + {y_3}}}{3}} \right)$

1. The Coordinates of the vertices of a triangle are (3, 1), (2, 3), and (-2, 2).  Find the coordinates of the centroid of the triangle ABC.
Applying the above formula, Centroid = $\left( {\displaystyle\frac{{3 + 2 - 2}}{3},\frac{{1 + 3 + 2}}{3}} \right)$ = (1, 2)

Incenter:  If $\left( {{x_{1,}}{y_1}} \right)$, $\left( {{x_{2,}}{y_2}} \right)$ and $\left( {{x_{3,}}{y_3}} \right)$ are the vertices of a triangle ABC such that BC = a, CA = b, AB = c, then the coordinates of its incenter = $\left( {\displaystyle\frac{{a{x_1} + b{x_2} + c{x_3}}}{{a + b + c}},\frac{{a{y_1} + b{y_2} + c{y_3}}}{{a + b + c}}} \right)$

2. Find the co-ordinates of the incenter of the triangle whose vertices are the points (4, -2), (5, 5),  and (-2, 4)
Let A =$\left( {{x_{1,}}{y_1}} \right)$ =  (4, -2), B = $\left( {{x_{2,}}{y_2}} \right)$ = (5, 5), and C = $\left( {{x_{3,}}{y_3}} \right)$ = (-2, 4)
Then a = BC = $\sqrt {{{\left( { - 2 - 5} \right)}^2} + {{\left( {4 - 5} \right)}^2}} = 5\sqrt 2$
b = AC = $\sqrt {{{\left( {4 + 2} \right)}^2} + {{\left( { - 2 - 4} \right)}^2}} = 6\sqrt 2$
c = AB = $\sqrt {{{\left( {5 - 4} \right)}^2} + {{\left( {5 + 2} \right)}^2}} = 5\sqrt 2$

The coordinates of the incenter of the $\Delta$ABC are $\left( {\displaystyle\frac{{a{x_1} + b{x_2} + c{x_3}}}{{a + b + c}},\frac{{a{y_1} + b{y_2} + c{y_3}}}{{a + b + c}}} \right)$ = $\left( {\displaystyle\frac{{5\sqrt 2 .4 + 6\sqrt {2.} 5 + 5\sqrt {2.} - 2}}{{5\sqrt 2 + 6\sqrt 2 + 5\sqrt 2 }},\frac{{5\sqrt 2 . - 2 + 6\sqrt {2.} 5 + 5\sqrt {2.} 4}}{{5\sqrt 2 + 6\sqrt 2 + 5\sqrt 2 }}} \right)$ = $\left( {\displaystyle\frac{{40\sqrt 2 }}{{16\sqrt 2 }},\frac{{40\sqrt 2 }}{{16\sqrt 2 }}} \right)$ = $\left( {\displaystyle\frac{5}{2},\frac{5}{2}} \right)$

Circum Center: If "O" is the circumcenter, all the vertices of the triangle ABC are equidistant from "O".  So to find the circumcenter we use the relationship OA = OB = OC.  Thhis gives two simultaneous linear equation and their solution provides the coordinates of circumcenter.

3. Find the coordinates of the circumcenter of the triangle whose vertices are (8, 6), (8, -2), and (2, -2)
Let S = (x, y) is the circumcenter. We know that SA = SB = SC
SA = SB $\Rightarrow S{A^2} = S{B^2}$
${\left( {x - 8} \right)^2} + {\left( {y - 6} \right)^2} = {\left( {x - 8} \right)^2} + {\left( {y + 2} \right)^2}$
$\Rightarrow$ 36 - 12y = 4 + 4y
$\Rightarrow$ y = 2

Again SB = SC $\Rightarrow S{B^2} = S{C^2}$
${\left( {x - 8} \right)^2} + {\left( {y + 6} \right)^2} = {\left( {x - 2} \right)^2} + {\left( {y + 2} \right)^2}$
$\Rightarrow$ - 12x+ 60 = 0
$\Rightarrow$ x = 5

The circumcenter S = (x, y) = (5, 2)

4. Find the circum center of the triangle (8,6), (6, 8), (-6, 8)
Sol: All the three points are equidistant from origin ie., $\sqrt {{8^2} + {6^2}}$
So circum center is origin. (0,0)

5. Find the orthocenter of the triangle (2,3), (3,2), (2,2)
Sol: (a,b), (b,a), (a, a) are the vertices then it is right angle triangle with right angle at (a, a). So orthocenter is (2,2)

6. Find the circumcenter of the triangle (0,0), (4, -4), (6, -3).
If the given vertices are $(0,0),({x_1},{y_1}),({x_2},{y_2})$ then
$\begin{array}{*{20}{c}} {2{x_1}}\\ {2{x_2}} \end{array}{\rm{ }}\begin{array}{*{20}{c}} {2{y_1}}\\ {2{y_2}} \end{array}{\rm{ }}\begin{array}{*{20}{c}} { - ({x_1}^2 + {y_1}^2)}\\ { - ({x_2}^2 + {y_2}^2)} \end{array}{\rm{ }}\begin{array}{*{20}{c}} {2{x_1}}\\ {2{x_2}} \end{array}$

Circumcenter = $\left( {\frac{{{\Delta _2}}}{{{\Delta _1}}},\frac{{{\Delta _3}}}{{{\Delta _1}}}} \right)$ where

${{\Delta _1}}$ = $2{x_1} \times 2{y_2}$

${\Delta _2} = 2{y_1} \times ( - ({x_2}^2 + {y_2}^2)) - 2{y_2} \times ( - ({x_1}^2 + {y_1}^2))$

${\Delta _3} = ( - ({x_1}^2 + {y_1}^2)) \times 2{x_2} - ( - ({x_1}^2 + {y_1}^2)) \times 2{x_1}$
$\begin{array}{*{20}{c}} 8\\ {12} \end{array}{\rm{ }}\begin{array}{*{20}{c}} { - 8}\\ { - 6} \end{array}{\rm{ }}\begin{array}{*{20}{c}} { - 32}\\ { - 45} \end{array}{\rm{ }}\begin{array}{*{20}{c}} 8\\ {12} \end{array}$
By cancelling 8 in top row and by 3 in bottom row we get  $\begin{array}{*{20}{c}} 1\\ 4 \end{array}{\rm{ }}\begin{array}{*{20}{c}} { - 1}\\ { - 2} \end{array}{\rm{ }}\begin{array}{*{20}{c}} { - 4}\\ { - 15} \end{array}{\rm{ }}\begin{array}{*{20}{c}} 1\\ 4 \end{array}$

${\Delta _1} = 2,{\Delta _2} = 7,{\Delta _3} = - 1$
Circumcenter = $\left( {\displaystyle\frac{7}{2},\frac{{ - 1}}{2}} \right)$

7. Find the circum center of the triangle formed by the points (-2,3), (2,-1), (4,0)
Shift origin to (0,0) by adding (2,-3)
Then the vertices become (0,0), (4, -4), (6, -3)
Solve similarly to above question.

8. Find the orthocenter of the triangle formed by Joining (0,0), (1, 2), (3, 8)
If the vertices are $(0,0),({x_1},{y_1}),({x_2},{y_2})$ then Orthocenter = $q({y_2} - {y_1}),q({x_2} - {x_1})$

( $q = \displaystyle\frac{{{x_1}{x_2} + {y_1}{y_2}}}{{{x_1}{y_2} - {x_2}{y_1}}}$

$q = \displaystyle\frac{{3 + 16}}{{8 - 6}} = \frac{{19}}{2}$

=$\left[ {\displaystyle\frac{{19}}{2}(6),\frac{{ - 19}}{2}(2)} \right] = (57, - 19)$

9. Find the image of (1,3) with respect to $x - y + 1 = 0$
Sol: If the image of $({x_1},{y_1})$ w.r.t ax + by + c = 0 in (x, y) then

$\displaystyle\frac{{x - {x_1}}}{a} = \frac{{y - {y_1}}}{b} = \frac{{ - 2(a{x_1} + b{y_1} + c)}}{{{a^2} + {b^2}}}$

$\Rightarrow \displaystyle\frac{{x - 1}}{1} = \frac{{y - 3}}{{ - 1}} = \frac{{ - 2(1 - 3 + 1)}}{{1 + 1}}$
Solving we get (x,y) = (2, 2)

10. Find the area of a triangle whose vertices are (1, 1), (5, 2) and (7, 4)
Solution:
Here $\left( {{x_{1,}}{y_1}} \right)$ = (1, 1) , $\left( {{x_{2,}}{y_2}} \right)$ = (5, 2) and $\left( {{x_{3,}}{y_3}} \right)$ = (7, 4)
Area of the triangle = $\displaystyle\frac{1}{2}\left[ {{x_1}\left( {{y_2} - {y_3}} \right) + {x_2}\left( {{y_3} - {y_1}} \right) + {x_3}\left( {{y_1} - {y_2}} \right)} \right]$
= $\displaystyle\frac{1}{2}\left( {1\;\left( {2\; - \;4} \right)\; + \;5\;\left( {4\; - \;1} \right)\; + \;7\;\left( {1\; - \;2} \right)} \right)$
= $\displaystyle\frac{1}{2}\left( { - \;2\; + \;15\; - \;7} \right)$
= $\displaystyle\frac{1}{2}\left( 6 \right)\; = \;3$ units

Short cut technique:  Finding triangle area by the above formula is cumbersome.  This problem can be solved easily by shifting the one of the coordinates to (0,0).
The area of the triangle whose coordinates are (0,0), $\left( {{x_{1,}}{y_1}} \right)$, $\left( {{x_{2,}}{y_2}} \right)$ is $\displaystyle\frac{1}{2}\left| {\left[ {{x_1}.{y_2} - {x_2}.{y_1}} \right]} \right|$

Now we can make the first coordinate (0,0), by adding ( -1, -1) to each of the co ordinates.  Now the shifted coordinates are (1-1,1-1), (5-1, 2-1), (7-1, 4-1) = (0,0), (4, 1), (6, 3)
Area of the triangle = $\displaystyle\frac{1}{2}\left| {\left[ {4.3 - 6.1} \right]} \right| = 3$ units

11. The area of the triangle whose vertices are (a, a), (a+1, a+1), (a+2, a) is [CAT - 2002]
Apply above short cut technique. Add (-a, -a) to all coordinates
So our new coordinates are (0, 0), (1, 1), (2, 0)
Area of the triangle = $\displaystyle\frac{1}{2}\left| {\left[ {1.0 - 1.2} \right]} \right|$=1 unit.

12. The intersection point of the lines represented by the equations $14x + 17y + 20 = 0$ and $99x + 102y + 105 = 0$
Sol: If a, b and c are in Arithmetic Progression, then $ax + by + c = 0$ passes through (1,- 2)

13. If l, m, n are in HP then find the fixed point through which the straight line $mnx + nly + lm = 0$ passes
Sol: (-1,2)

14. Find the intersection point of the lines $(a - b)x + (b - c)y + (c - a) = 0$ and $(p - q)x + (q - r)y + (r - p) = 0$
Sol: If a + b + c =0 then the straight line $ax + by + c = 0$ passes through (1, 1)