Question

21) If (x,y) be a point equidistant from the points (7,5) and (4,3)
(a) What is the distance between (x,y) and (7,5)?
Þ) What is the distance between (x,y) and (4,3)?
Show that 6x+4y=49?
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Answer 1

Answer:

Step-by-step explanation:

LetA(x,y),B(7,5) and C(4,3)

AB²=AC²

(x-7)²+(y-5)²=(x-4)²+(y-3)²

x²+49-14x+y²-10y+25=x²+16-8x+y²-6y+9

-14x-10y+74=-8x-6y+25

-14x+8y-10y+6y=25-74

-6x-4y=-49

6x+4y=49 Hence proved