Question

If A(-2, 5) , B(1, -3) and C(a, b) form an isosceles triangle with CB = CA, then​

Answer 1

Answer:

We have given A(-2,5) , B(1,-3) and C(a,b) form an isosceles triangle then the value of 6a-16b+19 = ? We know the distance formula between two points Now AB = sqrt[(-3 - (5))² + ((1) - (-2))²] AB = √73 BC = sqrt[(b - (-3))² + ((a) - 1)²] = sqrt[(b + 3))² + (a - 1)²] AC = sqrt[(b - 5))² + (a + 2)²] Since triangle is isosceles, and we know isosceles triangle have two sides are equal, so either AB=BC or AB=AC or BC=AC Let AB = BC √73 = sqrt[(b + 3))² + (a - 1)²] 73 = (b + 3))² + (a - 1)² ----------(i) and AB = AC √73 = sqrt[(b - 5))² + (a + 2)²] 73 = (b - 5))² + (a + 2)² -------------(ii) From equation (i) and (ii), we get => (b + 3))² + (a - 1)² = (b - 5))² + (a + 2)² Expanding the equation we get 6a - 16b + 19 = 0 Ans.Read more on Sarthaks.com - https://www.sarthaks.com/1071/if-a-2-5-b-1-3-and-c-a-b-form-an-isosceles-triangle-then-find-the-value-of-6a-16b-19

Answer 2

Answer:

Distance formula = $\sqrt{(x2-x1)^{2} + (y2-y1)^2 }$

given CB = CA

thus we can find the distance between them

Step-by-step explanation:

$\sqrt{(a-(-2))^{2} + (b-5)^2 } = \sqrt{(a-1)^{2} + (b-(-3))^2 }$

$\sqrt{(a+2))^{2} + (b-5)^2 } = \sqrt{(a-1)^{2} + (b+3))^2 }$

remove the square root :

${(a+2))^{2} + (b-5)^2 } = {(a-1)^{2} + (b+3))^2 }$

$(a + b)^2$ = $a^2+b^2-2ab$

$a^2+4+4a + b^2+25-10b = a^2+1-2a+b^2+9+6b$$a^2-a^2+4-9+25-1+4a+2a+b^2-b^2-10a-6a = 0$

$6a-16b+19 = 0$