Question

16. Under what condition the point (x, y) lie at equidistant from
(7,1) and (3,-5)?​

Answer 1

Answer: using distance formula

using distance formula→

using distance formula→ (x−7)

using distance formula→ (x−7) 2

using distance formula→ (x−7) 2 +(y−1)

using distance formula→ (x−7) 2 +(y−1) 2

using distance formula→ (x−7) 2 +(y−1) 2

using distance formula→ (x−7) 2 +(y−1) 2

using distance formula→ (x−7) 2 +(y−1) 2 =

using distance formula→ (x−7) 2 +(y−1) 2 = (x−3)

using distance formula→ (x−7) 2 +(y−1) 2 = (x−3) 2

using distance formula→ (x−7) 2 +(y−1) 2 = (x−3) 2 +(y−5)

using distance formula→ (x−7) 2 +(y−1) 2 = (x−3) 2 +(y−5) 2

using distance formula→ (x−7) 2 +(y−1) 2 = (x−3) 2 +(y−5) 2

using distance formula→ (x−7) 2 +(y−1) 2 = (x−3) 2 +(y−5) 2

using distance formula→ (x−7) 2 +(y−1) 2 = (x−3) 2 +(y−5) 2

using distance formula→ (x−7) 2 +(y−1) 2 = (x−3) 2 +(y−5) 2 ⇒−14x+49−2y+1=−6x+9−10y+25

using distance formula→ (x−7) 2 +(y−1) 2 = (x−3) 2 +(y−5) 2 ⇒−14x+49−2y+1=−6x+9−10y+25⇒8x−8y=16

using distance formula→ (x−7) 2 +(y−1) 2 = (x−3) 2 +(y−5) 2 ⇒−14x+49−2y+1=−6x+9−10y+25⇒8x−8y=16⇒x−y=2⇒y=x−2

Answer 2

Answer:

Answer

→ using distance formula

→

(x−7)

2

+(y−1)

2

=

(x−3)

2

+(y−5)

2

⇒−14x+49−2y+1=−6x+9−10y+25

⇒8x−8y=16

⇒x−y=2⇒y=x−2

$| |answer| |$

y=x−2