Question

If A (−3, 2), B(a, b) and C (1, 4) are the vertices of an isosceles triangle, prove that a + b = 1if AB =AC

Answer 1

Answer:

A(-3,2 ) ,B (a,b ) , c(1,4)

AB= AC

to prove : a+b=1

Solution :

Using distance formula ;

Distance Ab = $\sqrt{(X2-X1)^2+ (Y2-Y1)^2}$

AB= $\sqrt{(a+3)^2 + (b-2)^2}$

AB=$\sqrt{a^2 + 9+ 6a + b^2 - 4b +4 }$

Distance AC = $\sqrt{(X2-X1)^2+ (Y2-Y1)^2}$

AC= $\sqrt{(1+3)^2+(4-2)^2}$

AC= $\sqrt{4^2 + 2^2}$

AC = $\sqrt{18}$

AB =AC (Given )

$\sqrt{a^2 + 9+ 6a + b^2 - 4b +4 }$ =  $\sqrt{18}$

Squaring both sides ;

$a^{2} + b^2 +4+9 +6a- 4b$ = 18

$a^2+ b^2+ 13 -18 + 6a- 4b = 0$

Step-by-step explanation:

Answer 2

Answer:

A(-3,2),B (a,b ), c(1,4)

AB= AC

to prove : a+b=1

Solution :

Using distance formula ;

Distance Ab =

V(X2 – X1)2 + (Y2 – Y1)²

AB= V(a + 3)2 + (b – 2)²

AB=Va? + 9 + 6a + b2 – 4b + 4

Distance AC =

V(X2 - X1)2 + (Y2 – Y1)2

AC= V (1 + 3)2 + (4 – 2)²

AC= V42 + 22

AC = V18

AB =AC (Given)

Va? +9+ 6a + b² – 4b + 4 = V18

Squaring both sides ;

a? + b? + 4+9+ 6a – 4b = 18

a+ b2 + 13 – 18 + 6a 4b = 0