Note: If intersect at a point , then

Question 1: Find the point of intersection of the following pairs of lines :

Answer:

Therefore,

Therefore,

Therefore,

Question 2: Find the coordinates of the vertices of a triangle, the equations of whose sides are:

Answer:

i) Given equations:

… … … … … i)

… … … … … ii)

… … … … … iii)

First consider equations i) and ii):

Therefore,

Now consider equations ii) and iii):

Therefore,

Now consider equations iii) and i):

Therefore,

, ,

ii) Given equations:

… … … … … i)

… … … … … ii)

… … … … … iii)

First consider equations i) and ii):

Therefore,

Now consider equations ii) and iii):

Therefore,

Now consider equations iii) and i):

Therefore,

, ,

Question 3: Find the area of the triangle formed by the lines

, and

Answer:

i) Given equations:

… … … … … i)

… … … … … ii)

… … … … … iii)

First consider equations i) and ii):

Therefore,

Now consider equations ii) and iii):

Therefore,

Now consider equations iii) and i):

Therefore,

, ,

Area of triangle by the above vertices

ii) Given equations:

… … … … … i)

… … … … … ii)

… … … … … iii)

First consider equations i) and ii):

Therefore,

Now consider equations ii) and iii):

Therefore,

Now consider equations iii) and i):

Therefore,

Area of triangle by the above vertices

iii) Given equations:

… … … … … i)

… … … … … ii)

… … … … … iii)

First consider equations i) and ii):

Therefore,

Now consider equations ii) and iii):

Therefore,

Now consider equations iii) and i):

Therefore,

Area of triangle by the above vertices

Question 4: Find the equations of the medians of a triangle, the equations of whose sides are: ,

Answer:

Given equations:

… … … … … i)

… … … … … ii)

… … … … … iii)

Solving i) and ii) we get

Solving ii) and iii) we get

Solving iii) and i) we get

Midpoint of

Therefore the equation of median :

Midpoint of

Therefore the equation of median :

Midpoint of

Therefore the equation of median :

Hence the equations of the three medians are ,

Question 5: Prove that the lines form an equilateral triangle.

Answer:

Given equations:

… … … … … i)

… … … … … ii)

… … … … … iii)

Solving i) and ii) we get

Solving ii) and iii) we get

Solving iii) and i) we get

Therefore the is an equilateral triangle.

Question 6: Classify the following pairs of lines as co-incident, parallel or intersecting:

Answer:

We will transform the equation into the form

, the lines are intersecting.

We will transform the equation into the form

, and the intercepts are different, therefore the lines are parallel.

We will transform the equation into the form

, the lines are parallel. Their intercepts are also the same. Hence we can say that the lines are coincident line.

Question 7: Find the equation of the line joining the point to the point of intersection of the lines

Answer:

Therefore,

Therefore the equation of the line passing through is:

Question 8: Find the equation of a line passing through the point of intersection of the lines that has equal intercept on x-axis.

Answer:

Therefore,

If the intercepts of the line are equal, then the equation of the line is:

This line passes through , therefore

Question 9: Show that the area of the triangle formed by the lines is equal to are roots of the equation

Answer:

Now , therefore

Substituting we get

. Hence proved.

Question 10: If the straight line passes through the intersection of the lines , and and is parallel to , find .

Answer:

Therefore,

If passes through , then

… … … … … i)

is parallel to , the slope of both the lines is the same.

… … … … … ii)

Substituting ii) in i) we get

Question 11: Find the orthocenter of the triangle the equations of whose sides are

Answer:

Given equations:

… … … … … i)

… … … … … ii)

… … … … … iii)

Solving i) and ii) we get

Solving i) and iii) we get

Therefore equation of :

… … … … … iv)

Equation of :

… … … … … v)

Now solving equation iv) and v) i.e.

Therefore,

Question 12: Three sides of a triangle respectively. Find the equation of the altitude through the vertex .

Answer:

Therefore,

Therefore point of intersection

Therefore the equation of :

Question 13: Find the coordinates of the orthocenter of the, triangle whose vertices are .

Answer:

Therefore equation of :

… … … … … i)

Therefore equation of :

… … … … … ii)

Now solve i) and

Therefore,

Therefore the orthocenter is

Question 14: Find the coordinates of the incentre and centroid of the triangle whose sides have the equations .

Answer:

Given:

… … … … … i)

… … … … … ii)

… … … … … iii)

Solving i) and ii) i.e.

Therefore,

Therefore the point of intersection is

Solving ii) and iii) i.e.

Therefore,

Therefore the point of intersection is

Solving iii) and i) i.e.

Therefore,

Therefore the point of intersection is

Now we need to find the lengths of the lines .

Hence

Also

Therefore Centroid

Incenter

Question 15: Prove that the lines form a rhombus.

Answer:

Given:

… … … … … i)

… … … … … ii)

… … … … … iii)

… … … … … iv)

Solving i) and ii) we get

Solving ii) and iii) we get

Solving iii) and iv) we get

Solving iv) and v) we get

Now let’s find the lengths of the sides:

Since the four lines are equal, ABCD is a rhombus.

Question 16: Find the equation of the line passing through the intersection of the lines and parallel to line .

Answer:

Solving the above lines gives us the point of intersection as

Slope of line is

Therefore the slope of the required line is as they are parallel

Therefore the equation of the required line:

Question 17: Find the equation of the straight line passing through the intersection of the lines and perpendicular to line .

Answer:

Solving the above equations, we get the point of intersection .

Slope of the line is

Therefore the slope of the line perpendicular to line is

Therefore the equation of the required line: