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The points (K, 3), (2, -4) and (-K+1,-2)...

The points `(K, 3), (2, -4) and (-K+1,-2) ` are collinear. Find K. 

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To find the value of \( K \) such that the points \( (K, 3) \), \( (2, -4) \), and \( (-K + 1, -2) \) are collinear, we can use the concept of the section formula and the condition for collinearity. ### Step-by-step Solution: 1. **Understanding Collinearity**: Three points are collinear if the area of the triangle formed by them is zero. This can also be checked using the section formula. 2. **Assigning Points**: Let the points be: - \( A(K, 3) \) - \( B(2, -4) \) - \( C(-K + 1, -2) \) 3. **Using the Section Formula**: The point \( C \) divides the line segment \( AB \) in some ratio \( m:n \). We can find this ratio by comparing the y-coordinates. 4. **Setting Up the Equation**: The y-coordinate of point \( C \) is given by the section formula: \[ y_C = \frac{m \cdot y_B + n \cdot y_A}{m + n} \] Here, substituting the known values: \[ -2 = \frac{m \cdot (-4) + n \cdot 3}{m + n} \] 5. **Assuming a Ratio**: Let's assume the ratio is \( m:n = x:1 \). Then we can rewrite the equation as: \[ -2 = \frac{x \cdot (-4) + 1 \cdot 3}{x + 1} \] 6. **Cross-Multiplying**: Cross-multiplying gives: \[ -2(x + 1) = -4x + 3 \] Simplifying this: \[ -2x - 2 = -4x + 3 \] 7. **Rearranging the Equation**: Bringing all terms involving \( x \) to one side: \[ -2 + 3 = -4x + 2x \] This simplifies to: \[ 1 = -2x \] 8. **Solving for \( x \)**: Dividing both sides by -2: \[ x = -\frac{1}{2} \] 9. **Finding the Ratio**: The ratio \( m:n \) is \( -\frac{1}{2}:1 \) or \( 1:2 \) (after taking the absolute values). 10. **Using the x-coordinates**: Now we can apply the section formula for the x-coordinates: \[ x_C = \frac{m \cdot x_B + n \cdot x_A}{m + n} \] Substituting: \[ -K + 1 = \frac{1 \cdot 2 + 2 \cdot K}{1 + 2} \] This simplifies to: \[ -K + 1 = \frac{2 + 2K}{3} \] 11. **Cross-Multiplying Again**: Cross-multiplying gives: \[ 3(-K + 1) = 2 + 2K \] Expanding this: \[ -3K + 3 = 2 + 2K \] 12. **Rearranging**: Bringing all terms involving \( K \) to one side: \[ 3 - 2 = 2K + 3K \] This simplifies to: \[ 1 = 5K \] 13. **Solving for \( K \)**: Dividing both sides by 5 gives: \[ K = \frac{1}{5} \] ### Final Answer: The value of \( K \) is \( \frac{1}{5} \).
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ICSE-EQUATION OF A LINE-EXERCISE 14(B)
  1. (-2, 4), (4, 8), (10, 7) and (11,-5) are the vertices of a quadrilate...

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  2. Show that the points (a ,\ b+c),\ \ (b ,\ c+a) and (c ,\ a+b) are coll...

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  3. Find x, if the slope of the line joining (x, 2) and (8, -11) is - 3/4.

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  4. The side AB of an equilateral triangle ABC is parallel to the x-axis. ...

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  5. The side AB of a square ABCD is parallel to the x-axis. Find the slope...

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  6. The side AB of a square ABCD is parallel to the x-axis. Find the slope...

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  7. A (5, 4), B (-3, -2) and C (1, -8) are the vertices of a triangle ABC....

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  8. A (5, 4), B (-3, -2) and C (1, -8) are the vertices of a triangle ABC....

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  9. A (5, 4), B (-3, -2) and C (1, -8) are the vertices of a triangle ABC....

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  10. The slope of the side BC of a rectangle ABCD is 2/3. Find :  the slo...

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  11. The slope of the side BC of a rectangle ABCD is 2/3. Find : the slop...

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  12. Find the slope and the inclination of the line AB if : A = (-3, -2) ...

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  13. Find the slope and the inclination of the line AB if : A = (0, -sqrt...

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  14. Find the slope and the inclination of the line AB if : A = (-1, 2sq...

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  15. The points A(-3, 2), B(2, -1) and C(a, 4) are collinear. Find a.

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  16. The points (K, 3), (2, -4) and (-K+1,-2) are collinear. Find K.

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  17. Plot the points A (1, 1), B (4, 7) and C (4, 10) on a graph paper. Con...

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  18. Find the value(s) of k so that PQ will be parallel to RS. Given : P(...

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  19. Find the value(s) of k so that PQ will be parallel to RS. Given : P(...

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  20. Find the value of k so that PQ will be parallel to RS . P(5, -1), Q(6,...

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