*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:

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:

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 .

Answer:

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 .

Answer:

Please refer to the adjoining figure.

Therefore the slope of AD

Hence the equation of AD:

Therefore the slope of BE

Hence the equation of BE:

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:

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:

.

Hence the equation of mirror:

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

Answer:

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:

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 .

Answer:

Therefore the line perpendicular to the given line is:

… … … … … i)

Given points

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:

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 .

Answer:

.

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:

.

Mid point C lies on , therefore

… … … … … i)

Slope of mirror

Hence the image of the point on mirror

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

Answer:

.

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 .

Answer:

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 15: Prove that the lines 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.

Question 16: Find the equation of a line drawn perpendicular to the line through the point where it meets the y-axis.

Answer:

The point of intersection with y axis means

Therefore for

Hence the point of intersection is

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 17: The perpendicular from the origin to the line meets it at the point ). Find the values of .

Answer:

… … … … … i)

Therefore the slope of the line

Hence the slope of the line perpendicular to the given line

Therefore the equation of the required line:

… … … … … ii)

The point of intersection of line i) and ii) is given as .

Substituting in equation ii) we get

Therefore from i)

Hence,

Question 18: Find the equation of the right bisector of the given line: segment, joining the points .

Answer:

.

Therefore the slope of right bisector

Hence the equation of right bisector:

Question 19: The line through intersects the line at right angle. Find the value of .

Answer:

Question 20: Find the image of the point with respect to the line assuming the line to be a plane mirror.

Answer:

… … … … … i)

Slope of mirror

Therefore slope of a line perpendicular to mirror

Let the image be

Equation of the perpendicular line:

… … … … … ii)

satisfies both line i) and ii), therefore

… … … … … iii)

Also,

… … … … … iv)

Solving iii) and iv) we get

Therefore the image of the point with respect to line .

Question 21: Find the coordinates of the foot of the perpendicular from the point to the line .

Answer:

… … … … … i)

Slope of the line

Therefore slope of a line perpendicular

Equation of the perpendicular line:

… … … ii)

Solving i) and ii) we get the point of intersection which is which is also the foot of the perpendicular from

Question 22: Find the projection of the point on the line joining the points .

Answer:

.

Therefore the slope of perpendicular to

Equation of mirror:

… … … … … i)

Equation of perpendicular:

… … … … … ii)

Solving i) and ii) we get the point of intersection which is

Question 23: Find the equation of a line perpendicular to the line and at a distance of , units from the origin.

Answer:

Therefore the line perpendicular to the given line is:

… … … … … i)

Given that line i) is 3 units

Therefore ,

Therefore the equation of the required line:

Question 24: The line meets the x-axis at and y-axis at . The line through perpendicular to meets the x-axis and the line at respectively. If is the origin of coordinates, find the area of figure .

Answer:

… … … … … i)

The x and y intercept are respectively.

Line perpendicular to i) is … … … … … ii)

Line ii) passes through , substituting we get

Therefore equation of the perpendicular is … … … … … iii)

Line iii) meets x-axis at

Point of intersection of line iii) and i) is

Area of Area Area

Therefore Area of

Area of

Therefore the area of sq. units.

Question 25: Find the equation of the straight line which cuts off intercepts on x-axis twice that on y-axis and is at a unit distance from the origin.

Answer:

intercepts be

Therefore the equation of line is:

Changing this into normal form

( this is the length of perpendicular from origin)

sign is because the line could be on both sides of origin.

Therefore the equation of the required line is .

Question 26: The equations of perpendicular bisectors of the sides of a are respectively. If the point find the equation of the line.

Answer:

Given lines: … … … … … i) … … … … … ii)

Please refer to the adjoining figure.

Therefore equation of :

… … … … … iii)

Let the coordinate of be

Now is the mid point of , therefore

lies on equation i)

… … … … … iv)

is also on line iii). Therefore,

… … … … … v)

Solving iv) and v) we get

Similarly, we calculate

Therefore the slope of equation :

… … … … … vi)

be the mid point of . Therefore,

lies on line ii). Therefore,

… … … … … vii)

Also lies on vi)

… … … … … viii)

Solving vii) and viii) we get , and

Hence, equation of :