Answer :

Let p(x , 0) be the point which devides the line segment joining A(1, -5 ) and B(-4,5 ) in the ratio m : 1

Then the using section formula

(x , 0 ) = \(( \frac{m×(-4) + 1 × 1 }{m+1} , \frac{m×5+1×(-5)}{m+1} ) \)

\(\Rightarrow 0 = \frac{m×5+1×(-5)}{m+1} \)

\(\Rightarrow 0 = 5m - 5 \)

\(\Rightarrow m = 5 /5 = 1 \)

Hence the required ratio is 1 : 1

Since the ratio is 1 : 1 , so P is the mid point

\(\therefore \) x = \(\frac{m×(-4) + 1 × 1 }{m+1} \)

x = \(\frac{1×(-4) + 1 × 1 }{1+1} = \frac{-4+1}{2} = \frac{-3}{2} \)

\(\therefore ( \frac{-3}{2} , 0 )\) is the required point

- NCERT solutions for class 10 maths chapter 1 Real Numbers
- NCERT solutions for class 10 maths chapter 2 Polynomials
- NCERT solutions for class 10 maths chapter 3 Pair of linear equations in two variables
- NCERT solutions for class 10 maths chapter 4 Quadratic Equations
- NCERT solutions for class 10 maths chapter 5 Arithmetic Progressions
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