Distance Formula
Category : 9th Class
Let the coordinate of two points A and B be \[({{x}_{1}},{{y}_{1}})\] and \[({{x}_{2}},{{y}_{2}})\] respectively then the distance between two points A and B can be found in the following way:
Take points A and B draws AE perpendicular to BD.
Thus \[OC={{x}_{1}},AC={{y}_{1}}\]
\[OD={{x}_{2}},BD={{y}_{2}}\]
Now \[AE={{x}_{2}}-{{x}_{1}}\] and \[BE={{y}_{2}}-{{y}_{1}}\]
In right \[\Delta \text{AEB}\]
\[\text{A}{{\text{B}}^{\text{2}}}~~=\text{A}{{\text{E}}^{\text{2}}}+\text{B}{{\text{E}}^{\text{2}}}\]
\[\Rightarrow \]\[A{{B}^{2}}={{({{x}_{2}}-{{x}_{1}})}^{2}}+{{({{y}_{2}}-{{y}_{1}})}^{2}}\] \[\Rightarrow \]\[AB=\sqrt{{{({{x}_{2}}-{{x}_{1}})}^{2}}+{{({{y}_{2}}-{{y}_{1}})}^{2}}}\]
Important Points to Use Distance Formula
In various geometrical problems in which we need to use distance formula. Before using this formula the following points must keep in your mind:
1. If \[A=({{x}_{1}}-{{y}_{1}})\]and\[B=({{x}_{2}}-{{y}_{2}})\] then \[AB=\sqrt{{{({{x}_{2}}-{{x}_{1}})}^{2}}+{{({{y}_{2}}-{{y}_{1}})}^{2}}}\] or \[\sqrt{{{({{x}_{1}}-{{x}_{2}})}^{2}}+{{({{y}_{1}}-{{y}_{1}})}^{2}}}\]
2. The distance of point \[C(x,y)\] from origin is \[\sqrt{{{x}^{2}}+{{y}^{2}}}\].
3. To show three points P, Q. and R are collinear find PQ, QR and PR and then show that the greatest of these three is equal to the sum of other two.
4. Use the following geometrical facts:
(i) A triangle is equilateral if all sides are equal.
(ii) A quadrilateral is parallelogram if opposite sides are equal.
(iii) A quadrilateral is rectangle if opposite sides are equal and diagonals are equal.
(iv) A quadrilateral is rhombus if all four sides are equal.
The circumcentre of a circle whose vertices are (-1, 0), (7, -6) and (-2, 3) is.... .
(a) (5, 0)
(b) (3, -3)
(c) (-3, 3)
(d) (4, 3)
(e) None of these
Answer: (b)
Explanation:
Let the coordinate of circumcentre be \[p(x,y)\] and A = (-1, 0), B = (7, -6) and C = (-2, 3)
Since P is the circumcentre
\[\text{P}{{\text{A}}^{\text{2}}}=\text{P}{{\text{B}}^{\text{2}}}\]
\[\Rightarrow \]\[{{(x+1)}^{2}}+{{y}^{2}}={{(x-7)}^{2}}+{{(y+6)}^{2}}\]
\[\Rightarrow \]\[{{(x+1)}^{2}}-{{(x-7)}^{2}}={{(y+6)}^{2}}-{{y}^{2}}\]
\[\Rightarrow \]\[(x+1+x-7)(\bcancel{y}+1-\bcancel{y}+7)=(y+6+y)(\bcancel{y}+6-\bcancel{y})\]
\[\Rightarrow \]\[(2x-6).8=(2y+6).6\] \[\Rightarrow \]\[4(x-3)=12(y+3)\]
\[\Rightarrow \]\[4x-12=3y+9\] \[\Rightarrow \]\[4x-3y=21\] ...(i)
\[P{{A}^{2}}=P{{C}^{2}}\]
\[\Rightarrow \]\[{{(x+1)}^{2}}+{{y}^{2}}={{(x+2)}^{2}}+{{y}^{2}}\] \[\Rightarrow \]\[4x+6y+13=2x+1\]
\[\Rightarrow \]\[x+3y=-6\] ...(ii)
adding (i) and (ii), we get
\[\begin{align} & 4x-3y=21 \\ & \,\,\underline{x+3y=-6} \\ & \,\,5x\,\,\,\,\,\,\,\,\,\,=15 \\ \end{align}\]
\[x=3\]
Now put the value of x in (ii), we get
\[3+3y=-6\] \[\Rightarrow \] \[3y=-9\] \[y=-3\]
the coordinate of circumcentre is (3, - 3).
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