*Note: When slope and a point is given, the equation of the line is *

Question 1: Find the equation of the straight line passing through the point and having slope .

Answer:

Here and

Substituting in we get

Hence the equation of the straight line is

Question 2: Find the equation of the straight line passing through and inclined at an angle of with the x-axis.

Answer:

Here and

Substituting in we get

Hence the equation of the straight line is

Question 3: Find the equation of the line passing through with slope .

Answer:

Here and

Substituting in we get

Hence the equation of the straight line is

Question 4: Find the equation of the line passing through and inclined with x-axis at an angle of .

Answer:

Here

Also

Substituting in we get

Hence the equation of the straight line is

Question 5: Find the equation of the straight line which passes through the point and makes such an angle with the positive direction of x-axis whose sine is

Answer:

Given

We know,

Therefore and

Substituting in we get

Hence the equation of the straight line is

Question 6: Find the equation of the straight line passing through and making an angle of with the positive direction of y-axis

Answer:

and

Substituting in we get

Hence the equation of the straight line is

Question 7: Find the lines through the point making angle and with the x-axis. Also, find the lines parallel to them cutting the y-axis at a distance of units below the origin.

Answer:

Given

Also

Therefore the equation passing through and having and slope respectively are

Now let us find lines parallel to these lines but passing through

Therefore the equation passing through and having and slope respectively are

Question 8: Find the equations of the straight lines which cut off an intercept from the y-axis and are equally inclined to the axes.

Answer:

Given the lines are equally inclined to the axes.

Hence, their inclination with positive x-axis are and

Therefore for Line 1:

and for Line 2:

Hence the equation of Line 1:

And the equation for Line 2:

Question 9: Find the equation of the line which intercepts a length on the positive direction of the x-axis and is inclined at an angle of with the positive direction of y-axis.

Answer:

Here and

Therefore equation of the line is:

Question 10: Find the equation of the straight line which divides the join of the points and in the ratio and is also perpendicular to it.

Answer:

Given points and

The coordinate of the point that divides AB in the ratio of is

Therefore the equation of the required line:

Question 11: Prove that the perpendicular drawn from the point on the join of and divides it in the ratio .

Answer:

Given points and

is to . Therefore slope of

Therefore equation of is:

… … … … … i)

Equation of :

… … … … … ii)

Point of intersection will be found by solving the two equations.

Multiply i) by and ii) by and then subtract ii) from i) we get

Substituting in i) we get

Hence the point of intersection is

Therefore we see that divides AB in the ratio of

Question 12: Find the equations to the altitudes of the triangle whose angular points ate and .

Answer:

Given and .

Let , and are the altitudes as shown in the figure.

Therefore equation of :

Therefore equation of :

Therefore equation of :

Question 13: Find the equation of the right bisector of the line segment joining the points and .

Answer:

Given points and

Therefore the midpoint of

Slope of

Therefore the slope of the bisector

Hence the equation of the bisector:

Hence the equation of the right bisector is

Question 14: Find the equation of the line passing through the point and perpendicular to the line joining and .

Answer:

Given points and

Slope of

Therefore the slope of the perpendicular line

Hence the equation of the perpendicular which is passing through :

Hence the equation of the line is

Question 15: Find the equation of the right bisector of the line segment joining the points and .

Answer:

Given points and

Mid point of

Slope of

Therefore slope of bisector

Therefore the equation of the right bisector:

Hence the equation of the right bisector is