# 9th Class Mathematics Coordinate Geometry Distance Formula

Distance Formula

Category : 9th Class

### Distance Formula

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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