Question

the perpendicular bisector of the line segment joining the points A(2,3) and B(5,6) cuts Y-axis at

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Answer 1

Answer:

QUESTION:-

the perpendicular bisector of the line segment joining the points A(2,3) and B(5,6) cuts Y-axis at

ANSWER:-

Slope of AB = 6 - 3 / 5 - 2 = 3/3 = 1

Slope of line perpendicular to AB = -1

Let C be the midpoint of AB then ,

C = ( 5 + 2 / 2 , 6 + 3 / 2 )

= ( 7 / 2 , 9 / 2 )

Thus , equation of line passes through C with slope -1 is ;

(y - 9 / 2 ) = -1 . ( x - 7 / 2)

at x = 0 and y = 8.

[Thus , it cuts y - axis at 8. ]

Answer 2

Answer:

By midpoint formula:

$\tt{(x,y) = \frac{(x_{1} + x_{2})}{2} , \frac{(y_{1} + y_{2})}{2}}$

The midpoint of AB:

$\tt{C(x,y) = \frac{(2 + 5)}{2} , \frac{(3+6)}{2}}$

=> $\tt{C(x,y) = \frac{7}{2} , \frac{9}{2}}$

Perpendicular bisector of AB will lie on this point C.

Slope of AB = $\sf{\frac{rise}{run}}$ = 1

Slope of perpendicular bisector of AB = $\sf{-(slope^{-1})}$

We know that:

y = mx + b

Here, x and y are the points of the midpoint of AB

m is the slope of the bisector

b is the point where the slope intersects the y axis

=> (9/2) = (-1)(7/2) + b

=> b = 8

Thus, the line intersects y axis at 8

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