# Examples

Example-1

Find the coordinates of the point which divides the line segment joining the point (5, -2) and (9, 6) in the ratio 3:1.

Solution

Let the required point be (
and
The coordinates of the required point are (8, 4).

*x*,*y*). Then,Example-2

Find the ratio in which the point (2,

*y*) divides the join of (-4, 3) and (6, 3) and hence, find the value of*y*.Solution

Let the required ratio be
Then,
âˆ´ The required ratio is :1 i.e., 3:2
Also,

*k*:1.**When asked for ratio**

*m*:*n*, for convenience we take ratio as*m*/*n*:1 or*k*:1.Example-3

Two vertices of a triangle are (-1, 4) and (5, 2) and its centroid is (0, -3). Find the third vertex.

Solution

Let the third vertex be (

and
âˆ´
Hence, the third vertex of the triangle is (-4, -15).

*x*,*y*). Thenand

*x*= -4 and*y*= -15Example-4

Find the coordinates of incentre of the triangle whose vertices are (4, -2), (-2, 4) and (5, 5).

Solution

*x*,

*y*) be the coordinates of incentre of â–³ABC, Then,

*y*coordinate can be calculated.

Example-5

A line makes equal intercepts of length â€˜

*a*â€™ on the coordinate axes, intersecting the X-axis and Y-axis at A and B respectively. A circle is circumscribed about the triangle OAB, where O is the origin of the coordinate system. A tangent is drawn to this circle at the point O, the sum of the perpendicular distances of the vertices. A, B and O from this tangent is:Solution

AM + BN + OO =

Example-6

What is the area of the triangle whose vertices are (4, 4), (3, -16) and (3, -2)?

Solution

Let
Then area of the given triangle

Since area of a triangle cannot be negative, so area = 7 sq. units.

*x*_{1}= 4,*x*_{2}= 3,*x*_{3}= 3 and*y*_{1 }= 4,*y*_{2 }= -16,*y*_{3 }= -2.Example-7

What is the area of the triangle formed by the points (-5, 7), (-4, 5) and (1, -5)?

Solution

Let
Area of the triangle formed by given points

Hence, the given points are not forming any triangle, rather they are collinear.

*x*_{1 }= -5,*x*_{2 }= -4,*x*_{3}= 1 and*y*_{1}= 7,*y*_{2}= 5,*y*_{3}= -5.Example-8

Find the equation of the straight line which passes through (3, 4) and the sum of whose

*X*and*Y*intercepts is 14.Solution

Let the intercepts made by the line X-axis and Y-axis and Y-axis be a and (14 - a) respectively.

Then, its equation is
Since it passes through (3, 4), we have:

a = 6 and a = 7.

So, the required equation is: or ,
i.e., 4

Then, its equation is

So, the required equation is: or ,

*x*+ 3*y*= 24 or*x*+*y*= 7.Example-9

What is the equation of a line which passes through the point (-1, 3) and is perpendicular to the straight-line 5

*x*+ 3*y*+ 1 = 0?Solution

The equation of any line perpendicular to the line 5
Since the required line passes through the point (-1, 3), we have 3 Ã— (-1) - 5 Ã— 3 +
Hence, the required equation is 3

*x*+ 3*y*+ 1 = 0 is 3*x*- 5*y*+*K*= 0*K*= 0, or,*K*= 18*x*- 4*y*+ 18 = 0.Example-10

What is the point of intersection of the lines 2

*x*+ 3*y*= 5 and 3*x*- 4*y*= 10?Solution

To find out the point of intersection, we just need to solve the simultaneous equations.
2
3
Multiplying equation (1) by 3 and equation (2) by 2, 2
Doing equation (3) - equation (4) gives us:
Hence,
So point of intersection = ,

*x*+ 3*y*= 5 â€¦(1)*x*- 4*y*= 10 â€¦(2)*x*+ 3*y*= 5 becomes 6*x*+ 9*y*= 15 (equation 3) and 3*x*- 4*y*= 10 becomes 6*x*- 8*y*= 20 (equation 4).*y*= . This gives us*x*=