Question

7. Find the coordinates of the circumcentre of a triangle whose vertices are (-3,1),
(0.-2) and (1,3)​

Answer 1

Answer:

Say A,B,C are the vertices then

A=(−3,1), B=(0,−2), C=(1,3)

So, mid point of AB=(

2

−3+0

,

2

1−2

)=(−1.5,−.5)

Slope of AB=

0−(−.5)

−2−(−1.5)

=−1

Slope of the bisector is the negative reciprocal of the given slope.

So, the slope of the perpendicular bisector =1

Equation of AB with slope 1 and the coordinates (−1.5,−0.5) is,

[y–(−0.5)]=1[x–(−1.5)]

y−x=1 ......(1)

Similarly, for AC

Mid point of AC=(

2

−3+1

,

2

1+3

)=(−1,2)

Slope of AC=[

1−(−.5)

3−(−1.5)

]=3

Slope of the bisector is the negative reciprocal of the given slope.

So, the slope of the perpendicular bisector =−3

Equation of AC with slope −3 and the coordinates (−1,2) is,

(y–2)=−3(x+1)

y–2=−3x−3

y+3x=−1 ......(2)

On subtracting equation (1) from (2), We get

4x=−2 or x=−0.5

Substitute the value of x in equation (1), we get

y−(−0.5)=−1

y=1−0.5=0.5

So, the circumcentre is (−0.5,0.5).

Step-by-step explanation:

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