Question

In A (-1, 1), B(1, 3) and C(3, a) point and if AB = BC than find a in coordinate geometry​

Answer 1

$\sf{answer}$

formula required:-

Distance formula :

$d = \sqrt{(x_2 - x_1)^{2} + (y_2 - y_1)^{2} }$

given,

A (-1, 1), B(1, 3) and C(3, a).

to find:-

value of 'a'.

$\sf\underline{solution}$

first we find AB and next BC

in, AB:-

A = (-1,1)

let:- -1 = $\sf{x_1}$

1 = $\sf{y_1}$

B = (1,3)

1 = $\sf{x_2}$

3 = $\sf{y_2}$

.: AB = $\sqrt{(1-(-1))²+(3-1)²}$

= $\sqrt{(2)²+(2)²}$

= $\sqrt{4+4}$

= $\sqrt{8}$

= 2$\sqrt{2}$

in BC,

B = (1,3)

let

1 = $\sf{x_1}$

3 = $\sf{y_1}$

C = (3,a)

3 = $\sf{x_2}$

a = $\sf{y_2}$

.: BC = $\sqrt{(3-1)²+(a-3)²}$

= $\sqrt{(2)²+(a²-2(a)(3)+(-3)²}$

= $\sqrt{4+a²-6a+9}$

= √a²-6a+13

.: AB = BC

2√2 = √a²-6a+13

squaring on both sides

(2√2)² = (√a²-6a+13)²

4×2 = a²-6a+13

8 = a²-6a+13

a²-6a+13-8= 0

a²-6a+5 = 0

a²-5a-a+5 = 0

a(a-5)-1(a-5) = 0

(a-1)(a-5) = 0

.: a = 1,5

Answer 2

Answer:

Formulae used is distance formulae

√(x2-x1) ²+ ( y2-y1)²

Give that:

A(-1,1) B(1,3) C(3,a)

To Find :

the value of a

Solution:

First we have to find AB

in AB A=(-1,1)

So, -1=x1

then 1= y1

Now B = (1,3)

Let 1=x2

and 3=y2

.: AB = √ ( 1-(-1) )² + (3-1)²

= √(2)²+(2)²

= √ 4+4

= √8

= √2×4

=√2×√4

=2√2