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Mathematics Lab Activity-15 Class XI
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Mathematics Lab Activity-15 Class XI
Chapter - 9 Straight Lines
Activity - 15
Objective
To verify that the equation of a line passing through the point of intersection of two lines a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0 is of the form a1x + b1y + c1 = 0 + λ( a2x + b2y + c2 = 0)
Material Required
Cardboard, White
paper, pencil , adhesive.
Theory
Any equation of form ax + by + c = 0, where a and b are not zero simultaneously is called general linear equation or general equation of the line.
The equation of a
straight line passing through the point of intersection of the two straight
lines,
Procedure
1.Take a card board of convenient size and Paste a white paper on it .2. Draw two mutually perpendicular lines X’OX and Y’OY to represent x-axis and y- axis respectively.
3.Take a suitable but same scale for marking points on x and y-axis
4. Draw the graph of two intersecting lines and mark their point of intersection as (h, k) as shown in fig. 15.1
3x + 2y = 7 or 3x + 2y – 7 = 0
6. On plotting the graphs of two lines, we find that their point of intersections is (2, 1/2) as shown in fig. 15.2.
7. The equations of the line passing through the point of the intersection (2, 1/2) of the two given line is :
(3x - 2y - 5) + λ (3x + 2y - 7) = 0 At λ= 1, -1, 2, 1/2 ………..(1)
8. Putting λ = 1 in equation (1), we get
(3x - 2y - 5) + 1 (3x + 2y - 7) = 0
⇒ 6x - 12 = 0
Putting x = 2 we get
6 x 2 – 12 = 0
0 = 0
Therefore for λ = 1 , equation (1) is satisfied by the point of intersection of the two given lines.
9. Putting λ, in equation(1) we get
(3x - 2y - 5) - 1 (3x + 2y - 7) = 0
3x - 2y - 5 - 3x - 2y + 7 = 0
- 4y + 2 = 0
Putting y = 1/2 in this equation, we get
-4 x 1/2 + 2 = 0
-2 + 2 = 0
0 = 0
Therefore for λ = 1 , equation(1) is satisfied by the point of intersection of the two given lines.
Observations
1. For λ = 3, the equation of the line passing through the point of intersection of the two given lines i.e. equation (1) becomes,
(3x - 2y - 5) + 3 (3x + 2y - 7) = 0
Or 3x - 2y – 5 + 9x + 6y – 21 = 0
Or 12x + 4y – 26 = 0
Putting x = 2, y = 1/2 , we get
12 x 2 + 4 x 1/2 - 26 = 0
24 + 2 – 26 = 0
26 - 26 = 0
Therefore, for λ = 3 equation (1) is satisfied by the point of intersection of the two given lines.
2. Similarly we can prove that equation (1) is satisfied by the point of intersection of the two given lines for λ = 4, 5, 6 ……. . It verifies that the equation of a line passing through the point of intersection of the equations a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0 is of the form
a1x + b1y
+ c1 = 0 + λ( a2x + b2y + c2) = 0
Result
Equation of a line passing through the point of intersection of two lines a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0 is of the form a1x + b1y + c1 = 0 + λ( a2x + b2y + c2 = 0)
Applications
1. This activity
can be used to understand the result relating to the equation of a line passing
through the point of intersection of the two given lines.
2. This activity
also leads to the conclusion that an infinitely many lines pass through a fixed
point.
VIVA – VOICE
Q. 1.
What is the general equation of a straight line?
Ans The general equation of a straight line is ax + by + c = 0 Where a and b are not the zero simultaneously.
Q.2 what
is a slope of line?
Ans. Slope of a
line is defined as the tangent of the angle, which the line make with the
positive direction of x- axis.
Q.3 What is a condition that the two lines a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0 where
(i) parallel
to each other ?
(ii)
perpendicular to each other?
Ans (i)
(ii)
Q.4 How many lines can be drawn from the point of
intersection of two given straight lines?
Ans. Infinitely many lines can be drawn
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