Physics, asked by Anonymous, 9 months ago

solve it plz.... .........​

Attachments:

Answers

Answered by ItzRadhika
74

Then co- ordinate of C(0,y)

From A to C

A(6,5) , B(0,y)

By distance formula

AC= √(x1-x2)² +(y1-y2)²

AC = √(6-0)² +(5-y)²

AC= √6²+5²+y²-2×5×y

AC= √36+5²+y²-10y

AC= √y²-10y+61

From C to B

B(-4,3) , C(0,y)

By distance formula

BC= √(x1-x2)²+(y1-y2)²

BC= √(-4-0)² + (3-y)²

BC= √(-4)² + 3²+y²-2×3×y

BC= √16+9+y²-6y

BC = √y²-6y+25

Now, Point C is equidistant from AB

BC= AC

√y²-10y+61 = √y²-6y+25

Square both side

y²-10y+61= y²-6y+25

-10y+6y= -61+25

-4y= -36

y= 9

Point on y Axis is (0,9)

_______________________________

Add for your knowledge ⤵️⤵️

• Distance formula = √(x1-x2)²+ (y1-y2)²

• Any point on the x- axis is of the form(x,0)

• Any point on the y-axis is of the form (0,y)

_______________________________

Attachments:
Similar questions