Note: If is a line, then the equation perpendicular to this line is given by where .

If is a line, then the equation parallel to this line is given by where .

Question 1: Find the Equation of a line passing through the point and parallel to the line .

Answer:

Given line:

Therefore the line parallel to the given line is:

   … … … … … i)

This line passes through . Therefore substituting this in i) we get

Therefore the equation of the required line is

Question 2: Find the equation of a line passing through and perpendicular to the line

Answer:

Given line:

Comparing with we get

Therefore the line perpendicular to the given line is:

     … … … … … i)

This line passes through . Therefore substituting this in i) we get

Therefore the equation of the required line is

Question 3: Find the equation of the perpendicular bisector of the line joining the points and .

Answer:

Let and

Therefore the midpoint of is

Slope of

Therefore the slope of perpendicular bisector

Hence the equation of the perpendicular bisector:

Question 4: Find the equation of the altitude of a whose vertices are and .

Answer:

Please refer to the adjoining figure.

Slope of

Therefore the slope of AD

Hence the equation of AD:

Slope of

Therefore the slope of BE

Hence the equation of BE:

Slope of

Therefore the slope of CF

Hence the equation of CF:

Question 5: Find the equation of a line which is perpendicular to the line and which cuts off an intercept of units with the negative direction of y-axis

Answer:

Given line:

Comparing with we get

Therefore the line perpendicular to the given line is:

     … … … … … i)

This line passes through . Therefore substituting this in i) we get

Therefore the equation of the required line is

Question 6: If the image of the point with respect to a line mirror is find the equation of the mirror.

Answer:

Given and .

Therefore the midpoint of is

Slope of

Therefore the slope of mirror

Hence the equation of mirror:

Question 7: Find the equation of the straight line through the point and perpendicular to the line .

Answer:

Given line:

Comparing with we get

Therefore the line perpendicular to the given line is:

     … … … … … i)

This line passes through . Therefore substituting this in i) we get

Therefore the equation of the required line is

Question 8: Find the equation of the straight line perpendicular to and cutting off an intercept on the positive direction of the x-axis.

Answer:

Given line:

Comparing with we get

Therefore the line perpendicular to the given line is:

     … … … … … i)

This line passes through . Therefore substituting this in i) we get

Therefore the equation of the required line is

Question 9: Find the equation of the straight line perpendicular to and which passes through the mid-point of the line segment joining and .

Answer:

Given line:

Comparing with we get

Therefore the line perpendicular to the given line is:

     … … … … … i)

Given points and

Therefore the midpoint of is

This line passes through . Therefore substituting this in i) we get

Therefore the equation of the required line is

Question 10: Find the equation of the straight line line which has y-intercept equal to and is perpendicular to .

Answer:

Given line:

Comparing with we get

Therefore the line perpendicular to the given line is:

     … … … … … i)

This line passes through . Therefore substituting this in i) we get

Therefore the equation of the required line is

Question 11: Find the equation of the right bisector of the line segment joining the points and .

Answer:

Given and .

Therefore the midpoint of is

Slope of

Therefore the slope of right bisector

Hence the right bisector:

Question 12: Find the image of the point with respect to the line mirror .

Answer:

Given and .

Therefore the midpoint of is

Mid point C lies on , therefore

      … … … … … i)

Slope of

Slope of mirror

Therefore

Therefore

Hence the image of the point on mirror is

Question 13: If the image of the point with respect to the line mirror be find the equation of the mirror.

Answer:

Given and .

Therefore the midpoint of is

Slope of

Therefore the slope of mirror

Hence the equation of mirror:

Question 14: Find the equation to the straight line parallel to and passing through the middle point of the join of points and .

Answer:

Given line:

Therefore the line parallel to the given line is:

   … … … … … i)

Given and .

Therefore the midpoint of is

This line passes through . Therefore substituting this in i) we get

Therefore the equation of the required line is

Question 15: Prove that the lines and form a parallelogram

Answer:

Given lines:

Line 1:         

Line 2:   

Line 3:         

Line 4:   

Therefore slope of Line 1 and Line 3 is equal. Therefore the lines are parallel to each other.

Similarly,  slope of Line 2 and Line 4 is equal. Therefore the lines are parallel to each other.

Hence the four lines will form a parallogram.