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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 (xy). Then,
 
Description: 15432.png and Description: 15440.png
 
The coordinates of the required point are (8, 4).
 
 
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 k:1.
 
Then, Description: 15449.png
 
The required ratio is Description: 15461.png:1  i.e., 3:2
 
Also, Description: 15470.png
 

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 (xy). Then
Description: 15479.png and Description: 15488.png
 
 x = -4 and y = -15
 
Hence, the third vertex of the triangle is (-4, -15).
 
 
Example-4
Find the coordinates of incentre of the triangle whose vertices are (4, -2), (-2, 4) and (5, 5).
Solution
Description: 14244.png
 
Description: 14256.pngand Description: 14264.png
 
Let (xy) be the coordinates of incentre of △ABC, Then,
Description: 14272.png
 
Similarly, y coordinate can be calculated.
Description: 14341.png
 
So, the coordinates of the incentre are Description: 14349.png
 
 
Example-5
A line makes equal intercepts of length ‘a’ 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
Description: 15516.png
 
AM + BN + OO = Description: 14360.png
 
 
Example-6
What is the area of the triangle whose vertices are (4, 4), (3, -16) and (3, -2)?
Solution
Let x1 = 4, x2 = 3, x3 = 3 and y1 = 4, y2 = -16, y3 = -2.
 
Then area of the given triangle
Description: 14370.png
 
Description: 14383.png
 
Description: 14395.png
 
Since area of a triangle cannot be negative, so area = 7 sq. units.
 
 
Example-7
What is the area of the triangle formed by the points (-5, 7), (-4, 5) and (1, -5)?
Solution
Let x1 = -5, x2 = -4, x3 = 1 and y1 = 7, y2 = 5, y3 = -5.
 
Area of the triangle formed by given points
Description: 14402.png
Description: 14430.png
 
Hence, the given points are not forming any triangle, rather they are collinear.
 
 
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 Description: 14447.png
 
Since it passes through (3, 4), we have:
Description: 14506.png
 
a = 6 and a = 7.
So, the required equation is: Description: 14534.png or Description: 14541.png,
 
i.e., 4x + 3y = 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 5x + 3y + 1 = 0?
Solution
The equation of any line perpendicular to the line 5x + 3y + 1 = 0 is 3x - 5y + K = 0
 
Since the required line passes through the point (-1, 3), we have 3 × (-1) - 5 × 3 + K = 0, or, K = 18
 
Hence, the required equation is 3x - 4y + 18 = 0.
 
 
Example-10
What is the point of intersection of the lines 2x + 3y = 5 and 3x - 4y = 10?
Solution
To find out the point of intersection, we just need to solve the simultaneous equations.
 
2x + 3y = 5 …(1)
 
3x - 4y = 10 …(2)
 
Multiplying equation (1) by 3 and equation (2) by 2, 2x + 3y = 5 becomes 6x + 9y = 15 (equation 3) and 3x - 4y = 10 becomes 6x - 8y = 20 (equation 4).
 
Doing equation (3) - equation (4) gives us:
 
Description: 15571.png
Hence, y = Description: 14581.png. This gives us x = Description: 14570.png
 
So point of intersection = Description: 14590.pngDescription: 14606.png
 




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