Question

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

Answer 1

Given:-

• A,B and C are the vertices of an isosceles triangle.

• Coordinate of A is (-3,2)

• Coordinate of B is (a,b)

• Coordinate of C is (-1,4)

• AB = AC

To find:-

★ Using distance formula:-

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

$\:\implies\sf \sqrt{(-3 - a)^2 + (2 - b)^2}$

$\:\implies\sf \sqrt{9 + a^2 + 6a + 4 + b^2 - 4b}$

BC = $\sf \sqrt{(x_3 - x_2 )^2 + (y_3 - y_2 )^2}$

$\:\implies\sf \sqrt{(-1 - a)^2 + (4 - b)^2}$

$\:\implies\sf \sqrt{1 + a^2 + 2a + 16 + b^2 - 8b}$

★ According to question:-

$:\implies\sf Distance_{(AB)} = Distance_{(AC)}$

$\:\implies\sf \sqrt{9 + a^2 + 6a + 4 + b^2 - 4b} = \sqrt{1 + a^2 + 2a + 16 + b^2 - 8b}$

$:\implies\sf \sqrt{13 + a^2 + 6a + b^2 - 4b} = \sqrt{17 + a^2 + 2a + b^2 - 8b}$

$:\implies\sf 13 + \cancel{a^2} + 6a + \cancel{b^2} - 4b = 17 + \cancel{a^2} + 2a + \cancel{b^2} - 8b$

$:\implies\sf 13 + 6a - 4b = 17 + 2a - 8b$

$:\implies\sf 6a - 4b - 2a + 8b = 17 - 13$

$:\implies\sf 4a + 4b = 4$

$:\implies\sf \cancel{4}(a + b) = \cancel{4}$

$:\implies\sf {\underline{(a + b) = 1}}$

★ Hence solved!!

$\rule{200}3$

Answer 2

$\sf{\huge{\bold{\pink{\bigstar{\boxed{\boxed{Question}}}}}}}$

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

$\sf{\huge{\bold{\pink{\bigstar{\boxed{\boxed{Solution}}}}}}}$

$\sf{\bold{\green{\underline{\underline{Given}}}}}$

• triangle ABC is an isosceles triangle
• AB = AC
• coordinate of A = -3 , 2
• coordinate of B = a , b
• coordinate of C = -1 , 4

$\sf{\bold{\blue{\underline{\underline{To\:proof}}}}}$

• a + b = 1

$\sf{\bold{\purple{\underline{\underline{ explanation }}}}}$

• Distance btw. AB

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

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

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

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

• Distance btw. BC

$BC = \sf \sqrt{(x_3 - x_2 )^2 + (y_3 - y_2 )^2}$

$BC = \sf \sqrt{(-1 - a)^2 + (4 - b)^2}$

$BC = \sf \sqrt{1 + a^2 + 2a + 16 + b^2 - 8b}$

$\sf{\bold{\red{\boxed{ BC = \sf \sqrt{1 + a^2 + 2a + 16 + b^2 - 8b}}}}}$

• Compare

$\sf Distance{(AB)} = Distance{(AC)}$

$\sf \sqrt{9 + a^2 + 6a + 4 + b^2 - 4b} = \sqrt{1 + a^2 + 2a + 16 + b^2 - 8b}$

$\sf \sqrt{13 + a^2 + 6a + b^2 - 4b} = \sqrt{17 + a^2 + 2a + b^2 - 8b}$

$\sf 13 + \cancel{a^2} + 6a + \cancel{b^2} - 4b = 17 + \cancel{a^2} + 2a + \cancel{b^2} - 8b$

$\sf 13 + 6a - 4b = 17 + 2a - 8b$

$\sf 6a - 4b - 2a + 8b = 17 - 13$

$\sf 4a + 4b = 4$

$\sf \cancel{4}(a + b) = \cancel{4}$

$\sf {{(a + b) = 1}}$

$\sf{\bold{\red{\boxed{a + b = 1 }}}}$

............Hence Proved

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