Question

Show that the perpendicular bisectors of the sides of the triangle with vertices (7, 2).
(5,-2) and (-1, 0) are concurrent. Also find the co-ordinates of the point of concurrence.
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Answer 1

Answer:

first draw a triangle

A=(7,3)-(5,-2)=(6,0)✓

B=(7,2)-(-1,0)=(3,1)✓

C=(5,-2)-(-1,0)=(2,-1)✓

Answer 2

Perpendicular bisector are concurrent and the point is $(0,\frac{11}{3}).$

Step-by-step explanation:

Consider a triangle ABC.

To prove,

The perpendicular bisector of the sides of the given triangle having vertices (7,2),(5,-2),(-1,0) is concurrent.

We will first find the mid points of AB,BC,CA,where P,Q,R are the mid points of AB,BC,CA respectively.

$P(x,y)=[\frac{7-1}{2},\frac{2}{2}]$ = (3,1)

$Q(x_1,y_1)=[\frac{7+5}{2},\frac{2-2}{2}]$ = (6,0)

$R(x_2,y_2)=[\frac{5-1}{2},\frac {-2}{2}]$ = (2,-1)

Line equation formula,

$y-y_1=\frac{y_2-y_1}{x_2-x_1}\times (x-x_1)$

Now,the line equation for BP,

⇒$(y-1)=\frac{-2-1}{5-3}\times (x-3)$

⇒$(y-1)=\frac{-3}{2}\times (x-3)$

⇒$2(y-1)=-3\times(x-3)$

⇒2y-2 = -3x+9

⇒2y+3x=-9+2

⇒2y+3x=11........(1)

Now,equation for QC,

$(y-0)=\frac{0-0}{-1-1}\times (x-6)$

⇒y-0=0

⇒y=0

Also,equation for AR,

⇒$(y-2)=\frac{-1-2}{2-7}\times (x-7)$

⇒$y-2=\frac{-3}{-5}\times (x-7)$

⇒$y-2=\frac{3}{5}\times (x-7)$

⇒5(y-2)= 3(x-7)

⇒5y-10=3x-21

⇒5y-3x=-21+10

⇒5y-3x=-11......(3)

Now,the intersection point of line QC and BP,

y=0,

$x=\frac{11-2y}{3}=\frac{11}{3}$

Now,put by substituting these values in equation (3),

5y-3x=-11

$5(0)-3(\frac{11}{3})=-11$

Since LHS =RHS,

Hence,perpendicular bisector are concurrent and the point is$(0,\frac{11}{3}).$