Question

Q21. Find the coordinates of the circumcenter of the triangle whose vertices are A(5,-1) , B (-1,5) and C (6,6). Find its radius also.​

Answer 1

Step-by-step explanation:

Given :-

The triangle whose vertices are A(5,-1) , B (-1,5) and C (6,6).

To find:-

Find the coordinates of the circumcenter of the triangle whose vertices are A(5,-1) , B (-1,5) and C (6,6). Find its radius also. ?

Solution :-

Given points are :

A(5,-1) , B (-1,5) and C (6,6).

Let the coordinates of the circumcentre be P(x,y)

We know that

Circumcentre of a triangle is the point of intersection of all the three perpendicular bisectors of the sides of triangle.

So, the vertices of the triangle lie on the circumference of the circle.

PA = PB = PC

Finding Distance between P and A:-

Let (x1, y1)=(x,y)=>x1=x and y1=y

Let(x2, y2)=(5,-1)=>x2=5 and y2=-1

We know that

The distance between two points (x1, y1) and

(x2, y2) is √[(x2-x1)^2+(y2-y1)^2] units

=> PA = √[(5-x)^2+(-1-y)^2]

=> PA = √(25-10x+x^2+1+2y+y^2)

=> PA = √(x^2-10x+2y+y^2+26) units -----(1)

Finding Distance between P and B:-

Let (x1, y1)=(x,y)=>x1=x and y1=y

Let(x2, y2)=(-1,5)=>x2=-1 and y2=5

We know that

The distance between two points (x1, y1) and

(x2, y2) is √[(x2-x1)^2+(y2-y1)^2] units

=> PB = √[(-1-x)^2+(5-y)^2]

=> PB = √(x^2+1+2x+y^2+25y-10y)

=> PB = √(x^2+2x-10y+y^2+26) units -------(2)

Finding Distance between P and C:-

Let (x1, y1)=(x,y)=>x1=x and y1=y

Let(x2, y2)=(6,6)=>x2=6 and y2=6

We know that

The distance between two points (x1, y1) and

(x2, y2) is √[(x2-x1)^2+(y2-y1)^2] units

=> PC = √[(6-x)^2+(6-y)^2]

=> PC = √(36-12x+x^2+36-12y+y^2)

=> PC= √(x^2-12x-12y+y^2+72) units -------(3)

On taking PA = PB

√(x^2-10x+2y+y^2+26) = √(x^2+2x-10y+y^2+26)

On squaring both sides

=>x^2-10x+2y+y^2+26 = x^2+2x-10y+y^2+26

=> -10x+2y = 2x-10y

=> 2y+10y = 2x+10x

=> 12y = 12x

=>x =y -------------(4)

On taking PB = PC

√(x^2+2x-10y+y^2+26) = √(x^2-12x-12y+y^2+72)

On squaring both sides

=>x^2+2x-10y+y^2+26= x^2-12x-12y+y^2+72

=> 2x-10y +26 = -12x-12y+72

=> 2x+12x +26 = 10y-12y+72

=> 14x+26 = -2y+72

=> 14x+2y = 72-26

=> 14x+2y = 46

=> 14x+2x = 46

=> 16x = 46

=> x = 46/16

=> x = 23/8

=>x = 23/6

=> y = 23/8

(x,y) = (23/8,23/8)

Now , PA

=> PA =√[(5-(23/8))^2+(-1-(23/8))^2]

=>PA = √[{40-23)/8}^2+{-8-23)/8}^2]

=> PA =√[(17/8)^2+(-31/8)^2]

=> PA =√[(289/64)+(961/64)]

=> PA =√[(289+961)/64]

=> PA =√(1250/64)

=>PA =√[(2×25×25)/(64)]

=>PA =25√2/8

=> PA = (25/8)√2 units

PA = PB=PC = (25/8)√2 units

or we know that

√2 = 1.414 then

Circumradius = (25×1.414)/8

=> 35.35/8

=> 4.1875

=> 4.2 units (approximately)

Answer:-

The circumcentre of the given trianagle formed by the given vertices is (23/8,23/8)

The circum radius of the given triangle = (25/8)√2 units or 4.2 units (approximately)

Used formulae:-

Circumference:-

The concurrent point of the bisectors of the sides of a triangle.

The distance between two points (x1, y1) and

(x2, y2) is √[(x2-x1)^2+(y2-y1)^2] units

Answer 2

$\large\underline{\sf{Solution-}}$

We know,

Circumcenter is the point of intersection of perpendicular bisectors of any two sides of a triangle.

Let assume that m and n are the perpendicular bisectors of sides AB and AC intersecting each other at 0.

Let D and F

are the midpoints of AB and AC respectively.

To find equation of m.

m is perpendicular bisector of AB.

So, m is perpendicular to AB.

$\rm :\longmapsto\:Slope \: of \: AB = \dfrac{5 + 1}{ - 1 - 5}$

$\rm :\longmapsto\:Slope \: of \: AB = \dfrac{6}{ - 6}$

$\rm :\longmapsto\:Slope of AB = - 1$

So,

$\rm :\longmapsto\:Slope \: of \: m = 1$

Now, D is the midpoint of AB.

So, by using midpoint Formula, we have

$\rm :\longmapsto\:Coordinates \: of \:D = \bigg(\dfrac{5 - 1}{2} , \: \dfrac{ - 1 + 5}{2} \bigg)$

$\rm :\longmapsto\:Coordinates \: of \: D= \bigg(\dfrac{4}{2} , \: \dfrac{4}{2} \bigg)$

$\rm :\longmapsto\:Coordinates \: of \: D= (2,2)$

Hence,

The equation of line m, which passes through the point (2, 2) and having slope 1, is

$\rm :\longmapsto\:y - 2= 1(x - 2)$

$\rm :\longmapsto\:y - 2= x - 2$

$\rm :\longmapsto\:y= x - - - - - (1)$

To find equation of line n

Since, n is perpendicular bisector of AC

So, n is perpendicular to AC.

$\rm :\longmapsto\:Slope \: of \: AC = \dfrac{6 + 1}{6 - 5}$

$\rm :\longmapsto\:Slope \: of \: AC = \dfrac{7}{1}$

$\rm :\longmapsto\:Slope \: of \: AC = 7$

So,

$\rm :\longmapsto\:Slope \: of \: n = - \: \dfrac{1}{7}$

Now, F is the midpoint of AC.

So, coordinates of F, using midpoint Formula is

$\rm :\longmapsto\:Coordinates \: of \: F = \bigg(\dfrac{5 + 6}{2} , \: \dfrac{6 - 1}{2} \bigg)$

$\rm :\longmapsto\:Coordinates \: of \: F = \bigg(\dfrac{11}{2} , \: \dfrac{5}{2} \bigg)$

Hence,

Equation of line n is

$\rm :\longmapsto\:y - \dfrac{5}{2} = - \dfrac{1}{7}(x - \dfrac{11}{2})$

$\rm :\longmapsto\:7y - \dfrac{35}{2} = - y + \dfrac{11}{2} \: \: \: \{ \: using \: (1) \}$

$\rm :\longmapsto\:8y = \dfrac{35}{2} + \dfrac{11}{2}$

$\rm :\longmapsto\:8y = \dfrac{35 + 11}{2}$

$\rm :\longmapsto\:8y = \dfrac{46}{2}$

$\rm :\longmapsto\:8y =23$

$\rm :\longmapsto\:y = \dfrac{23}{8}$

So,

$\rm :\longmapsto\:x = \dfrac{23}{8}$

Hence,

Coordinates of Circum- center, O is

$\rm :\longmapsto\:Coordinates \: of \:O = \bigg(\dfrac{23}{8} , \: \dfrac{23}{8} \bigg)$

Now, we find the radius of circle,

$\rm :\longmapsto\:r = OC = \sqrt{ {(6 - \dfrac{23}{8} )}^{2} + {(6 - \dfrac{23}{8} )}^{2} }$

$\rm \: \: \: = \sqrt{ {(\dfrac{48 - 23}{8} )}^{2} + {(\dfrac{48 - 23}{8} )}^{2} }$

$\rm \: \: \: = \sqrt{ {(\dfrac{25}{8} )}^{2} + {(\dfrac{25}{8} )}^{2} }$

$\rm \: \: \: = \sqrt{ {(\dfrac{25}{8})}^{2} \times 2 }$

$\rm \: = \: \: \dfrac{25}{8} \sqrt{2} \: units$

Formula used :-

1. Slope of a line :-

Let us consider a line segment passes through the point (a, b) and (c, d), then slope is given by

$\rm :\longmapsto\:m = \dfrac{d - b}{c - a}$

2. Two lines having slope m and M are perpendicular iff Mm = - 1

3. Slope point form of a line

Let us consider a line passes through the point (a, b) having slope m, then equation is given by

y - b = m ( x - a ).

4. Distance Formula :-

${\underline{\boxed{\rm{\quad Distance = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \quad}}}}$

5. Midpoint Formula :-

$\boxed{ \quad\sf \:( x, y) = \bigg(\dfrac{x_1+x_2}{2} , \dfrac{y_1+y_2}{2} \bigg) \quad}$