solve it plz.... .........
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Answers
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)
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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)
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