Question

find the locus of third vertex of Right angled triangle the ends of the hypothenuse are (4, 0) (0, 4
)​

Answer 1

Answer:-

$\textbf{Let the Poinťś be } = \: P(x,y), \: A(4 ,0) , \: B(0,4)$

$\text{As A And B are-ordinates of AB}$

$⇒AB \sqrt{(4 - 0 {)}^{2} + (0 - 4 {)}^{2} }$

$⇒ \sqrt{32}$

$⇒PA = \sqrt{(4 - x {)}^{2} + (0 - y {)}^{2} }$

$⇒ \sqrt{(4 - x {)}^{2} + {y}^{2} }$

$⇒PB = \sqrt{(4 - y {)}^{2} + (0 - x {)}^{2} }$

$⇒ \sqrt{ {x}^{2} + (4 - y {)}^{2} }$

$\text{As ∆PAB is right- angled}$

$⇒∠P = 90°$

$∴(PA {)}^{2} + (PB {)}^{2} = (AB {)}^{2}$

$⇒(4 - x {)}^{2} + {y}^{2} + {x}^{2} + (4 - y {)}^{2} = 32$

$⇒16 + {x}^{2} - \frac{16x}{2} + {y}^{2} + {x}^{2} + {y}^{2} + 16 - 8y = 32$

$⇒ {2x}^{2} + {2y}^{2} - 8x - 8y = 0$

$⇒ {x}^{2} + {y}^{2} - 4x - 4y = 0$

Hence,

$\text{Required locus is \: } (x - 2 {)}^{2} + (y - 2 {)}^{2} = 8$

$\\ \\ \\ \\ \sf \colorbox{lightgreen} {\red★ANSWER ᵇʸɴᴀᴡᴀʙﷻ}$