Math, asked by mishratanya392, 6 months ago

Find the point which lies on the perpendicular bisector of the line segment joining the points A (1, 5) and

B (4, 6) cuts the y-axis.​

Answers

Answered by Arighnach
2

Answer:

0,13)

(0,-13)

(0,12)

(13,0)

Answer :

a

Solution :

Firstly, we plot the points of the segment on the paper and join them.

We know that , the Perpendicular bisector of the line segment AB bisect the segment AB, i.e., perpendicular bisector of line segment AB passes through the mid - point of AB.

∴ Mid - [point of AB (1+42,5+62)

⇒P=(52,112)

[∴ mid point of line segment passes through the points (x1+y1) and (x2+y2)=(x1+x22,y1+y22)]

Now , we draw a straigth line on paper passes through the mid - point P . We see that the perpendicular bisector cuts the Y- axis at the point (0,13). Hence, the required point is (0,13).

Step-by-step explanation:

Alternate Method

We know that , the equation of line which passes through the points (x1,y1)and(x2,y2) is

(y−y1)=y2−y1x2−x1(x−x1) ...(i)

Here, x1=,y1=5andx2=4,y2=6

So , the equation of line segment joining the points A(1,5) and B (4,6) is

(y−5)=6−54−1(x−1)

⇒(y−5)=13(x−1)

⇒3y−15=x−1

⇒3y=x−14⇒y=13x−143 ...(ii)

∴ Slope of line segment , m1=13

If two lines are perpendicular to each other , then the relation between its slopes is

m1⋅m2=−1 ...(iii)

where, m1 = Slope of line 1

and = Slope of line 2

Also, we know that the perpendicular bisector of the line segment is perpendicular on the line segment. Let slope of line segment is m2.

From Eq. (iii),

m1⋅m2=13⋅m2=−1

⇒m2=−3

Also we know that perpendicular bisector is passes through the mid- point of line segment.

∴ Mid - point of line segment =(1+42,5+62)=(52,112)

Equation of perpendicular bisector , which has slope (-3) and passes through the point (52,112), is

(y−112)=−(−3)(x−52)

[Since , equation of line passes through the point (x1,y1) and having slope m (y−y1)=m(x−x1)]

⇒(2y−11=−2(2x−5)

⇒2y−11=−6x+15

⇒6x+2y=26

⇒3x+y=13 ...(iv)

If perpendicular bisector cuts the y- axis , then put x=0 in Eq. (iv),

3×0+y=13⇒y=13

So , the required point is (0,13).

mark me as  brainliest

Similar questions