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Show that the Straight Lines Given by (2 + K) X + (1 + K) Y = 5 + 7k for Different Values of K Pass Through a Fixed Point. Also, Find that Point.

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Question

Show that the straight lines given by (2 + k) x + (1 + k) y = 5 + 7k for different values of k pass through a fixed point. Also, find that point.

Answer in Brief
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Solution

The given straight line (2 + k)x + (1 + k)y = 5 + 7k can be written in the following way:
2x + y − 5 + k (x + y − 7) = 0
This line is of the form L1 + kL2 = 0, which passes through the intersection of the lines
L1 = 0 and L2 = 0, i.e. 2x + y − 5 = 0 and x + y − 7 = 0.
Solving 2x + y − 5 = 0 and x + y − 7 = 0, we get (−2, 9), which is the fixed point.

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Equations of Line in Different Forms - Equation of Family of Lines Passing Through the Point of Intersection of Two Lines
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Chapter 23: The straight lines - Exercise 23.19 [Page 131]

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R.D. Sharma Mathematics [English] Class 11
Chapter 23 The straight lines
Exercise 23.19 | Q 7 | Page 131

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