Co-ordinate Geometry

1.0Distance Formula

The distance between two points $\left({\mathrm{x}}_1,{\mathrm{y}}_1\right)$ and $\left({\mathrm{x}}_2,{\mathrm{y}}_2\right)$ in a rectangular coordinate system is equal to $\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}$. Proof: X'OX and Y'OY are the rectangular coordinate axes. $\mathrm{P}\left({\mathrm{x}}_1,{\mathrm{y}}_1\right)$ and $\mathrm{Q}\left({\mathrm{x}}_2,{\mathrm{y}}_2\right)$ are the given points. We draw PA and QB perpendiculars on the x -axis; PC and QD perpendiculars on the $y$-axis.

2.0Distance of a point from origin

The distance of a point $(x,y)$ from origin is $\sqrt{x^2+y^2}$. Proof: Let us take a point $\mathrm{P}(\mathrm{x},\mathrm{y})$ in the given plane of axes ${\mathrm{X}}^{\prime }\mathrm{OX}$ and ${\mathrm{Y}}^{\prime }\mathrm{OY}$ as shown in the figure. Here, the point $P(x,y)$ is in the first quadrant but it can be taken anywhere in all the four quadrants. We have to find the distance $OP$, i.e., the distance of the point $P$ from the origin 0 . From the point P , draw $\mathrm{PM}\perp \mathrm{OX}$ and $\mathrm{PL}\perp \mathrm{OY}$.

Test for Geometrical Figures:

(a) For an isosceles triangle : Prove that two sides are equal. (b) For an equilateral triangle : Prove that three sides are equal. (c) For a right-angled triangle : Prove that the sum of the squares of two sides is equal to the square of the third side. (d) For a square : Prove that all sides are equal and diagonals are equal. (e) For a rhombus : Prove that all sides are equal and diagonals are not equal. (f) For a rectangle : Prove that the opposite sides are equal and diagonals are also equal. (g) For a parallelogram : Prove that the opposite sides are equal in length and diagonals are not equal.

Numerical Ability 1 Find the distance between the points $\left(a\cos 35^{\circ },0\right)$ and $\left(0,a\cos 55^{\circ }\right)$ Solution: Required distance $=\sqrt{{\left({\mathrm{x}}_2-{\mathrm{x}}_1\right)}^2+{\left({\mathrm{y}}_2-{\mathrm{y}}_1\right)}^2}$ $=\sqrt{{\left(0-\mathrm{a}\cos 35^{\circ }\right)}^2+{\left(\mathrm{a}\cos 55^{\circ }-0\right)}^2}$ $=\sqrt{{\mathrm{a}}^2{\cos }^{2}35^{\circ }+{\mathrm{a}}^2{\cos }^{2}55^{\circ }}$ $=\mathrm{a}\sqrt{{\cos }^{2}35^{\circ }+{\sin }^{2}35^{\circ }}=\mathrm{a}$ units

• If two opposite vertices of a square are ( ${\mathrm{x}}_1,{\mathrm{y}}_1$ ) and ( ${\mathrm{x}}_2,{\mathrm{y}}_2$ ), then its area is $\frac{1}{2}\left|{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2\right|$.

Numerical Ability 2 Find points on $x$-axis which are at a distance of 5 units from the point $A(-1,4)$. Solution: Let the point on x -axis be $\mathrm{P}(\mathrm{x},0)$. Distance $=$ PA $=5$ units $\Rightarrow {\mathrm{PA}}^2=25$ sq. units $\Rightarrow (\mathrm{x}+1{)}^2+(0-4{)}^2=25$ $\Rightarrow x^2+2x+1+16=25$ $\Rightarrow x^2+2x+17=25$ $\Rightarrow x^2+2\mathrm{x}-8=0$ $\Rightarrow {\mathrm{x}}^2+4\mathrm{x}-2\mathrm{x}-8=0$ $\Rightarrow \mathrm{x}(\mathrm{x}+4)-2(\mathrm{x}+4)=0$ The point of inter section of altitudes of a triangle is called orthocentre. $\Rightarrow (x-2)(x+4)=0$ $\mathrm{x}=2,-4$ Required point on $x$-axis are $(2,0)$ and $(-4,0)$ Verification: $PA=\sqrt{(2+1{)}^2+(0-4{)}^2}=\sqrt{9+16}$ $=\sqrt{25}=5$ and also, $\mathrm{PA}=\sqrt{(-4+1{)}^2+(0-4{)}^2}$ $=\sqrt{9+16}$ $=\sqrt{25}=5$

Some examples of simplifications, how to solve fast:

• $\sqrt{34^2+34^2}=34\sqrt{2}$
• $\sqrt{15^2+18^2}=3\sqrt{5^2+6^2}=3\sqrt{31}$

Building Concept 1 The vertices of a triangle are $(-2,0),(2,3)$ and $(1,-3)$. Is the triangle equilateral, isosceles or scalene? Explanation: We denote the given point $(-2,0),(2,3)$ and $(1,-3)$ by $A,B$ and $C$ respectively then

Collinearity of three points

Let $A,B$ and $C$ are three given points. Point $A,B$ and $C$ will be collinear, if the sum of lengths of any two line-segments is equal to the length of the third line-segment.

In the adjoining fig. there are three-point $\mathrm{A},\mathrm{B}$ and C . Three points $\mathrm{A},\mathrm{B}$ and C are collinear if and only if (i) $\mathrm{AB}+\mathrm{BC}=\mathrm{AC}$ (ii) $\mathrm{AB}+\mathrm{AC}=\mathrm{BC}$ (iii) $\mathrm{AC}+\mathrm{BC}=\mathrm{AB}$

Numerical Ability 3 Prove that the points $A(1,1),B(-2,7)$ and $C(3-3)$ are

$$\begin{gathered} \mathrm{AB}=\sqrt{(-2-1{)}^2+(7-1{)}^2} \\ =\sqrt{(-3{)}^2+6^2}=\sqrt{45}=3\sqrt{5}\text{ units, } \\ \mathrm{BC}=\sqrt{(3+2{)}^2+(-3-7{)}^2} \\ =\sqrt{5^2+(-10{)}^2}=\sqrt{125}=5\sqrt{5}\text{ units, } \\ \mathrm{AC}=\sqrt{(3-1{)}^2+(-3-1{)}^2} \\ =\sqrt{2^2+(-4{)}^2}=\sqrt{20}=2\sqrt{5}\text{ units. } \\ \therefore \mathrm{AB}+\mathrm{AC}=(3\sqrt{5}+2\sqrt{5})\text{ units }=5\sqrt{5}\text{ units }=\mathrm{BC} \end{gathered}$$ Thus, $\mathrm{AB}+\mathrm{AC}=\mathrm{BC}$ Hence, the given points A, B, C are collinear.

• The co-ordinate of the point which divides the line segment joining the points ( ${\mathrm{x}}_1,{\mathrm{y}}_1$ ) and ( ${\mathrm{x}}_2,{\mathrm{y}}_2$ ) externally in the ratio $\mathrm{m}:\mathrm{n}$ are $\left(\frac{{\mathrm{mx}}_2-{\mathrm{nx}}_1}{\mathrm{m}-\mathrm{n}},\frac{{\mathrm{my}}_2-{\mathrm{ny}}_1}{\mathrm{m}-\mathrm{n}}\right)$
• In a right angled triangle, the circum-centre is the mid point of hypotenuse and the orthocentre is the point where right angle is formed.

3.0Section formula

Coordinates of the point, dividing the line-segment joining the points $\left({\mathrm{x}}_1,{\mathrm{y}}_1\right)$ and $\left({\mathrm{x}}_2,{\mathrm{y}}_2\right)$ internally in the ratio $m_1:m_2$ are given by $\left(\frac{m_1{x}_2+m_2{x}_1}{m_1+m_2},\frac{m_1{y}_2+m_2{y}_1}{m_1+m_2}\right)$

Proof : Let $\mathrm{P}(\mathrm{x},\mathrm{y})$ be the point dividing the line-segment joining $\mathrm{A}\left({\mathrm{x}}_1,{\mathrm{y}}_1\right)$ and $\mathrm{B}\left({\mathrm{x}}_2,{\mathrm{y}}_2\right)$ internally in the ratio $m_1:m_2$. We draw the perpendiculars $AL,BM$ and $PQ$ on the $x$-axis from the points $A,B$ and $P$ respectively. $L,M$ and $Q$ are the points on the $x$-axis where these perpendiculars meet the $x$-axis.

We draw $\mathrm{AC}\perp \mathrm{PQ}$ and $\mathrm{PD}\perp \mathrm{BM}$. Here $\mathrm{AC}\Vert \mathrm{x}$-axis and $\mathrm{PD}\Vert \mathrm{x}$-axis. $\Rightarrow \mathrm{AC}\Vert \mathrm{PD}$ ( $\because$ AC and PD both || x -axis) $\Rightarrow \angle \mathrm{PAC}=\angle \mathrm{BPD}$

Thus, in $△\mathrm{ACP}$ and $△\mathrm{PDB}$, we have $\angle \mathrm{PAC}=\angle \mathrm{BPD}$ and $\angle \mathrm{ACP}=\angle \mathrm{PDB}=90^{\circ }$. Then by AA similarity criterion, $△\mathrm{ACP}\sim \Delta \mathrm{PDB}$

$$\begin{gathered} \Rightarrow \frac{\mathrm{AC}}{\mathrm{PD}}=\frac{\mathrm{PC}}{\mathrm{BD}}=\frac{\mathrm{AP}}{\mathrm{PB}}=\frac{{\mathrm{m}}_1}{{\mathrm{m}}_2} \\ \Rightarrow \frac{\mathrm{AC}}{\mathrm{PD}}=\frac{{\mathrm{m}}_1}{{\mathrm{m}}_2} \end{gathered}$$

and $\frac{\mathrm{PC}}{\mathrm{BD}}=\frac{{\mathrm{m}}_1}{{\mathrm{m}}_2}$ $AC=LQ=OQ-OL=\left(x-x_1\right)$ $\mathrm{PD}=\mathrm{QM}=\mathrm{OM}-\mathrm{OQ}=\left({\mathrm{x}}_2-\mathrm{x}\right)$

Putting in (i), we get $\frac{\left(x-x_1\right)}{\left(x_2-x\right)}=\frac{m_1}{m_2}$ $\Rightarrow {\mathrm{m}}_2\mathrm{x}-{\mathrm{m}}_2{\mathrm{x}}_1={\mathrm{m}}_1{\mathrm{x}}_2-{\mathrm{m}}_1\mathrm{x}$ $\Rightarrow m_1\mathrm{x}+{\mathrm{m}}_2\mathrm{x}={\mathrm{m}}_1{\mathrm{x}}_2+{\mathrm{m}}_2{\mathrm{x}}_1$ $\Rightarrow \left(m_1+m_2\right)x=m_1{x}_2+m_2{x}_1$ $\Rightarrow \mathrm{x}=\frac{{\mathrm{m}}_1{\mathrm{x}}_2+{\mathrm{m}}_2{\mathrm{x}}_1}{{\mathrm{m}}_1+{\mathrm{m}}_2}$ Now, $\mathrm{PC}=\mathrm{PQ}-\mathrm{CQ}=\mathrm{PQ}-\mathrm{AL}=\left(\mathrm{y}-{\mathrm{y}}_1\right)$ $BD=BM-DM=BM-PQ=\left(y_2-y\right)$ Putting in (ii), we get $$\begin{gathered} \frac{y-y_1}{y_2-y}=\frac{m_1}{m_2} \\ \Rightarrow y=\frac{m_1{y}_2+m_2{y}_1}{m_1+m_2} \end{gathered}$$ Therefore, the coordinates of the point P are $\left(\frac{m_1{x}_2+m_2{x}_1}{m_1+m_2},\frac{m_1{y}_2+m_2{y}_1}{m_1+m_2}\right)$ Remark: To remember the section formula, the diagram given below is helpful

• In an equilateral triangle, the centroid, incentre, orthocentre and circumcentre coincide.

Point dividing a line segment in the ratio \mathrm{k}: 1 If $\mathrm{P}\left(\frac{m_1{x}_2+m_2{x}_1}{m_1+m_2},\frac{m_1{y}_2+m_2{y}_1}{m_1+m_2}\right)$ divides the line-segment, joining $A\left(x_1,y_1\right)$ and $B\left(x_2,y_2\right)$ internally in the ratio $m_1:m_2$ we can express it as below $P\left(\frac{\frac{m_1}{m_2}{x}_2+x_1}{\frac{m_1}{m_2}+1},\frac{\frac{m_1}{m_2}{y}_2+y_1}{\frac{m_1}{m_2}+1}\right)$ [By dividing the numerator and the denominator by ${\mathrm{m}}_2$ ]

Putting $\frac{m_1}{m_2}=k$, the ratio becomes $\mathrm{k}:1$ and the coordinates of P are expressed in the form $\mathrm{P}\left(\frac{{\mathrm{kx}}_2+{\mathrm{x}}_1}{\mathrm{k}+1},\frac{{\mathrm{ky}}_2+{\mathrm{y}}_1}{\mathrm{k}+1}\right)$.

Therefore, the coordinates of the point P , which divides the line-segment joining $\mathrm{A}\left({\mathrm{x}}_1,{\mathrm{y}}_1\right)$ and $B\left(x_2,y_2\right)$ internally in the ratio $\mathrm{k}:1$, are given by $\left(\frac{{\mathrm{kx}}_2+{\mathrm{x}}_1}{\mathrm{k}+1},\frac{{\mathrm{ky}}_2+{\mathrm{y}}_1}{\mathrm{k}+1}\right)$

• In an isosceles triangle, the centroid, incentre, orthocentre and circumcentre are collinear.

4.0Mid-point formula

Coordinates of the mid-point of the line-segment joining $\left({\mathrm{x}}_1,{\mathrm{y}}_1\right)$ and $\left({\mathrm{x}}_2,{\mathrm{y}}_2\right)$ are $\left(\frac{{\mathrm{x}}_1+{\mathrm{x}}_2}{2},\frac{{\mathrm{y}}_1+{\mathrm{y}}_2}{2}\right)$ The mid-point $\mathrm{M}(\mathrm{x},\mathrm{y})$ of the line-segment joining $\mathrm{A}\left({\mathrm{x}}_1,{\mathrm{y}}_1\right)$ and $\mathrm{B}\left({\mathrm{x}}_2,{\mathrm{y}}_2\right)$ divides the linesegment $AB$ in the ratio $1:1$. Putting $m_1=m_2=1$ in the section formula, we get the coordinates of the mid-point as $\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)$.

Collinearity of three points

Three given points $A\left(x_1,y_1\right),B\left(x_2,y_2\right),C\left(x_3,y_3\right)$ are said to be collinear if one of them must divide the line segment joining the other two points in the same ratio.

• Three points are called non-collinear if one of them divides the line segment joining the other two points in different ratios.

Numerical Ability 4 Show by section formula that the points $(3,-2),(5,2)$ and $(8,8)$ are collinear. Solution: Let the point $\mathrm{B}(5,2)$ divides the line joining $\mathrm{A}(3,-2)$ and $\mathrm{C}(8,8)$ in the ratio $\mathrm{m}:\mathrm{n}$, Then by section formula, $x=\frac{\left({\mathrm{mx}}_2+{\mathrm{nx}}_1\right)}{\mathrm{m}+\mathrm{n}}$ $$\begin{gathered} \therefore 5=\frac{(m\times 8+n\times 3)}{m+n} \\ \therefore 5=\frac{(8m+3n)}{m+n} \\ 5m+5n=8m+3n \\ 2n=3m \\ \therefore \frac{m}{n}=\frac{2}{3} \end{gathered}$$ By section formula, $y=\frac{\left({\mathrm{my}}_2+{\mathrm{ny}}_1\right)}{\mathrm{m}+\mathrm{n}}$ $$\begin{gathered} \therefore 2=\frac{(m\times 8+n\times -2)}{m+n} \\ \therefore 2=\frac{(8m-2n)}{m+n} \\ 2m+2n=8m-2n \\ 6m=4n \\ \frac{m}{n}=\frac{4}{6}=\frac{2}{3} \end{gathered}$$ Here ratios are same So, the points are collinear.

Numerical Ability 5 Find the co-ordinates of the points which divide the line segment joining $\mathbf{A}(-3,3)$ and $B(3,9)$ into four equal parts. Solution:

Numerical Ability 6 If the point $C(-1,2)$ divides the line segment $AB$ in the ratio $3:4$, where the coordinates of A are $(2,5)$, find the coordinates of $B$. Solution:

• While finding ratio in the questions of section formula, use K : 1. It will be easier rather than taking $\mathrm{m}:\mathrm{n}$.

Numerical Ability 7 Find the ratio in which the line segment joining the points $(1,-7)$ and $(6,4)$ is divided by x -axis. Solution: Let $\mathrm{C}(\mathrm{x},0)$ divides AB in the ratio $\mathrm{k}:1$. By section formula, the coordinates of $C$ are given by $C\left(\frac{6k+1}{k+1},\frac{4k-7}{k+1}\right)$ But C $(x,0)=C\left(\frac{6k+1}{k+1},\frac{4k-7}{k+1}\right)$ $\Rightarrow \frac{4\mathrm{k}-7}{\mathrm{k}+1}=0$ $\Rightarrow 4\mathrm{k}-7=0$ $\Rightarrow \mathrm{k}=\frac{7}{4}$ i.e., the x -axis divides AB in the ratio $7:4$.

Numerical Ability 8 Find the value of $m$ for which coordinates $(3,5),(m,6)$ and $\left(\frac{1}{2},\frac{15}{2}\right)$ are collinear. Solution Let $\mathrm{P}(\mathrm{m},6)$ divides the line segment AB joining $\mathrm{A}(3,5),\mathrm{B}\left(\frac{1}{2},\frac{15}{2}\right)$ in the ratio $\mathrm{k}:1$.

If $\left({\mathrm{x}}_1,{\mathrm{y}}_1\right),\left({\mathrm{x}}_2,{\mathrm{y}}_2\right)$, $\left({\mathrm{x}}_3,{\mathrm{y}}_2\right)\dots \left({\mathrm{x}}_{\mathrm{n}},{\mathrm{y}}_{\mathrm{n}}\right)$ are the vertices of a polygon of $n$ sides, its area $=$ $\frac{1}{2}|\left(x_1{y}_2-x_2{y}_1\right)+\left(x_2{y}_3-x_3{y}_2\right)$

5.0Centroid of a triangle

• The point of intersection of internal bisectors of the angles of the triangle is called incentre. The co-ordinates of the incentre of the triangle ABC whose co-ordinates $\mathrm{A}\left({\mathrm{x}}_1\mathrm{y}1\right),\mathrm{B}\left({\mathrm{x}}_2{\mathrm{y}}_2\right)$ and $\mathrm{C}(\mathrm{x}3,\mathrm{y}3)$, and $\mathrm{BC}=\mathrm{a},\mathrm{AC}=\mathrm{b}$ and $\mathrm{AB}=\mathrm{c}$, are $\frac{a{x}_1+b{x}_2+c{x}_3}{a+b+c},\frac{a{y}_1+b{y}_2+c{y}_3}{a+b+c}$
• Euler line: The circumcentre 0 , the centroid $G$ and the orthocentre $H$ of a triangle are collinear, the line on which they lie is called Euler line. Also G divides HO in the ratio $2:1$ i.e. $\frac{\mathrm{OG}}{\mathrm{GH}}=\frac{1}{2}$
  

Numerical Ability 9 Find the centroid of $△ABC$ whose vertices are $A(-3,0),B(5,-2)$ and $C(-8,5)$. Solution: Here, $\left({\mathrm{x}}_1=-3,{\mathrm{y}}_1=0\right),\left({\mathrm{x}}_2=5,{\mathrm{y}}_2=-2\right)$ and $\left({\mathrm{x}}_3=-8,{\mathrm{y}}_3=5\right)$. Let $G(x,y)$ be the centroid of $△ABC$. Then, $\mathrm{x}=\frac{1}{3}\left({\mathrm{x}}_1+{\mathrm{x}}_2+{\mathrm{x}}_3\right)=\frac{1}{3}(-3+5-8)=-2$, $\mathrm{y}=\frac{1}{3}\left({\mathrm{y}}_1+{\mathrm{y}}_2+{\mathrm{y}}_3\right)=\frac{1}{3}(0-2+5)=1$. Hence, the centroid of $△ABC$ is $G(-2,1)$.

Numerical Ability 10 Two vertices of a $△ABC$ are given by $A(6,4)$ and $B(-2,2)$, and its centroid is $G(3,4)$. Find the coordinates of the third vertex $C$ of $△ABC$. Solution: Two vertices of $△ABC$ are $A(6,4)$ and $B(-2,2)$. Let the third vertex be $C(a,b)$. Then, the coordinates of its centroid are $G\left(\frac{6-2+a}{3},\frac{4+2+b}{3}\right)$,i.e., $G\left(\frac{4+a}{3},\frac{6+b}{3}\right)$. But, it is given that the centroid is $\mathrm{G}(3,4)$. $$\begin{gathered} \therefore \frac{4+a}{3}=3\text{ and }\frac{6+b}{3}=4 \\ \Rightarrow a=5\text{ and }b=6 \end{gathered}$$ Hence, the third vertex of $△ABC$ is $C(5,6)$.

• In your previous classes, you have learnt to find the area of a triangle in terms of its base and corresponding altitude as shown below: Area of triangle $=\frac{1}{2}\times$ base $\times$ altitude .
• In case, we know the lengths of the three sides of a triangle, then the area of the triangle can be obtained by using the Heron's formula.

6.0Area of a triangle

In this section, we will find the area of a triangle when the coordinates of its three vertices are given. The lengths of the three sides can be obtained by using distance formula but we will not prefer the use of Heron's formula. Sometime, the lengths of the sides are obtained as irrational numbers and the application of Heron's formula becomes tedious. Let us develop some easier way to find the area of a triangle when the coordinates of its vertices are given.

Let $\mathrm{A}\left({\mathrm{x}}_1,{\mathrm{y}}_1\right),\mathrm{B}\left({\mathrm{x}}_2,{\mathrm{y}}_2\right)$ and $\mathrm{C}\left({\mathrm{x}}_3,{\mathrm{y}}_3\right)$ be the given three points. Through A draw $AQ\perp OX$, through $B$ draw $\mathrm{BP}\perp \mathrm{OX}$ and through C draw $\mathrm{CR}\perp \mathrm{OX}$.

Condition of collinearity of three points

The given points $\mathrm{A}\left({\mathrm{x}}_1,{\mathrm{y}}_1\right),\mathrm{B}\left({\mathrm{x}}_2,{\mathrm{y}}_2\right)$ and $\mathrm{C}\left({\mathrm{x}}_3,{\mathrm{y}}_3\right)$ will be collinear if the area of the triangle formed by them must be zero because triangle cannot be formed. $\Rightarrow$ area of $△\mathrm{ABC}=0$ $\Rightarrow \frac{1}{2}\left|{\mathrm{x}}_1\left({\mathrm{y}}_2-{\mathrm{y}}_3\right)+{\mathrm{x}}_2\left({\mathrm{y}}_3-{\mathrm{y}}_1\right)+{\mathrm{x}}_3\left({\mathrm{y}}_1-{\mathrm{y}}_2\right)\right|=0$ $\Rightarrow \left|{\mathrm{x}}_1\left({\mathrm{y}}_2-{\mathrm{y}}_3\right)+{\mathrm{x}}_2\left({\mathrm{y}}_3-{\mathrm{y}}_1\right)+{\mathrm{x}}_3\left({\mathrm{y}}_1-{\mathrm{y}}_2\right)\right|=0$ is the required condition for three points to be collinear.

Numerical Ability 11 Find the area of the triangle whose vertices are $A(2,7),B(3,-1)$ and $C(-5,6)$. Solution Let $A(2,7),B(3,-1)$ and $C(-5,6)$ be the vertices of the given $△ABC$. Then, $\left(x_1=2,y_1=7\right),\left(x_2=3,y_2=-1\right)$ and ( $x_3=-5,y_3=6)$. Area of $\Delta ABC=\frac{1}{2}\left|\left\{x_1\left(y_2-y_3\right)+x_2\left(y_3-y_1\right)+x_3\left(y_1-y_2\right)\right\}\right|$ $=\frac{1}{2}|2(-1-6)+3(6-7)-5(7+1)|$ $=\frac{1}{2}|-14-3-40|=\frac{1}{2}|-57|=\frac{57}{2}$ $=28.5$ sq. units.

Building Concept 2 Show that the points $A(a,b+c),B(b,c+a)$ and $C(c,a+b)$ are collinear. Explanation Let $\mathrm{A}(\mathrm{a},\mathrm{b}+\mathrm{c}),\mathrm{B}(\mathrm{b},\mathrm{c}+\mathrm{a})$ and $\mathrm{C}(\mathrm{c},\mathrm{a}+\mathrm{b})$ be the given points. Then, $\left({\mathrm{x}}_1=\mathrm{a},{\mathrm{y}}_1=\mathrm{b}+\mathrm{c}\right)$; $\left({\mathrm{x}}_2=\mathrm{b},{\mathrm{y}}_2=\mathrm{c}+\mathrm{a}\right)$; and ( ${\mathrm{x}}_3=\mathrm{c},{\mathrm{y}}_3=\mathrm{a}+\mathrm{b})$. $$\begin{gathered} \therefore x_1\left(y_2-y_3\right)+x_2\left(y_3-y_1\right)+x_3\left(y_1-y_2\right) \\ =a(c+a-a-b)+b(a+b-b-c)+c(b+c-c-a) \\ =a(c-b)+b(a-c)+c(b-a) \\ \Rightarrow ac-ab+ba-bc+cb-ca=0 \end{gathered}$$ Hence, the given points are collinear.

7.0Mind Map