1.

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

Answer»

\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


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