Question

33.
A triangle whose vertex are (2, 3), (4, 5) and
(-2, 11) distance between circumcentre and vertex
(4, 5) is-
(1) 2
(2) 4/5
(3) 2/5
(4) 4​

Answer 1

Answer:

2√5

Step-by-step explanation:

A=(2,3)

B=(4,5)

C=(-2,11)

First we need to find the circumcentre

slope of AB=$\frac{5-3}{4-2}=\frac{2}{2}=1$

slope of CA=$\frac{11-3}{-2-2}=\frac{8}{-4}=-2$

The midpoints of AB and CA are (3,4) and(0,7) respectively

let midpoint of AB be P and midpoint of CA be Q and circumcentre be O

so P=(3,4) and Q=(0,7)

since PO is the perpendicular bisector of AB.therefore slope of PO is -1

similarly QO is perpendicular bisector of CA.therefore slope of QO is 2

equation of line PO is y-4=-1(x-3)

equation of line QO is y-7=2(x-0)

solving both the equations we get x=0 and y=7

now using the distance formula and calculating the distance from vertex(4,5)

$D=\sqrt{(7-5)^{2}+(0-4)^{2} }  } =\sqrt{4+16} =2\sqrt{5}$

Answer 2

Answer:

2 answer............