Question

Find the equation of straight line which is perpendicular to line 2x+y-8=0 and passing through mid-point of line segment joining (0,7) and (8,7). ​

Answer 1

Answer:

Given line : 3x+2y−8=0

2y=−3x+8

y=(  

2

−3

​

)x+4

Here, slope (m  

1

​

)=  

2

−3

​

 

Now, the co-ordinates of the mid-point of the line segment joining the points (5,−2) and (2,2) will be

(  

7

(5+2)

​

,  

7

(−2+2)

​

)=(  

2

7

​

,0)

Let's consider the slope of the perpendicular to the given line be m  

2

​

 

Then,

m  

1

​

×m  

2

​

=−1

(  

2

−3

​

)×m  

2

​

=−1

m  

2

​

=  

3

2

​

 

So, the equation of the line with slope m  

2

​

 and passing through (  

2

7

​

,0) will be

y−0=(  

3

2

​

)(x−  

2

7

​

)

3y=2x−7

2x−3y−7=0

Thus, the required line equation is 2x−3y−7=0