Question

if the linex-4y-6=0​

Answer 1

Step-by-step explanation:

Given, the line equation : x−4y−6=0 .....(i)

Co-ordinates of P are (1,3)

Let the co-ordinates of Q be (x,y)

Now, the slope of the given line is

4y=x−6

y=(

4

1

)x−

4

6

slope (m)=

4

1

So, the slope of PQ will be

m

−1

[Astheproductofslopesofperpendicularlinesis−1]

Slope of

PQ=

(

4

1

)

−1

=−4

Now, the equation of line PQ will be

y−3=(−4)(x−1)

y−3=−4x+4

4x+y=7 ... (ii)

On solving equation (i) and (ii), we get the coordinates of M

Multiplying (ii) by 4 and adding with (i), we get

x−4y−6=0

16x+4y=28

17x=34

x=

17

34

=2

Putting the value of x in (i)

2−4y−6=0

−4−4y=0

4y=−4

y=−1

So, the co-ordinates of M are (2,−1)

But, M is the mid-point of line segment PQ

(2,−1)=

2

(x+1)

,

2

(y+3)

2

(x+1)

=2

x+1=4

x=3

And,

2

(y+3)

=−1

y+3=−2

y=−5

Thus, the co-ordinates of Q are (3,−5)

solution

Answer 2

Step-by-step explanation:

Given, the line equation : x−4y−6=0 .....(i)

Co-ordinates of P are (1,3)

Let the co-ordinates of Q be (x,y)

Now, the slope of the given line is

4y=x−6

$y = ( \frac{1}{4} )x - \frac{6}{4}$

$slope \: (m) = \frac{1}{4}$

So, the slope of PQ will be

m

−1

[Astheproductofslopesofperpendicularlinesis−1]

Slope of

PQ=

(

4

1

)

−1

=−4

Now, the equation of line PQ will be

y−3=(−4)(x−1)

y−3=−4x+4

4x+y=7 ... (ii)

On solving equation (i) and (ii), we get the coordinates of M

Multiplying (ii) by 4 and adding with (i), we get

x−4y−6=0

16x+4y=28

17x=34

x=

17

34

=2

Putting the value of x in (i)

2−4y−6=0

−4−4y=0

4y=−4

y=−1

So, the co-ordinates of M are (2,−1)

But, M is the mid-point of line segment PQ

(2,−1)=

2

(x+1)

,

2

(y+3)

2

(x+1)

=2

x+1=4

x=3

And,

2

(y+3)

=−1

y+3=−2

y=−5

Thus, the co-ordinates of Q are (3,−5)