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The two line segment joining (-2,7), (-5...

The two line segment joining `(-2,7), (-5,-3) and (-8,-13),(1,17)` cut each other at

A

only one point

B

no point

C

infinite number of points

D

none of these

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To find the point of intersection of the two line segments joining the points (-2, 7) to (-5, -3) and (-8, -13) to (1, 17), we will derive the equations of both lines and analyze their relationship. ### Step 1: Find the equation of the first line segment The first line segment connects the points A(-2, 7) and B(-5, -3). Using the two-point form of the line equation: \[ \frac{y - y_1}{y_2 - y_1} = \frac{x - x_1}{x_2 - x_1} \] Where \((x_1, y_1) = (-2, 7)\) and \((x_2, y_2) = (-5, -3)\). Substituting the values: \[ \frac{y - 7}{-3 - 7} = \frac{x + 2}{-5 + 2} \] This simplifies to: \[ \frac{y - 7}{-10} = \frac{x + 2}{-3} \] Cross-multiplying gives: \[ 3(y - 7) = -10(x + 2) \] Expanding both sides: \[ 3y - 21 = -10x - 20 \] Rearranging the equation: \[ 10x + 3y - 1 = 0 \quad \text{(Equation 1)} \] ### Step 2: Find the equation of the second line segment The second line segment connects the points C(-8, -13) and D(1, 17). Using the same two-point form: \[ \frac{y - y_1}{y_2 - y_1} = \frac{x - x_1}{x_2 - x_1} \] Where \((x_1, y_1) = (-8, -13)\) and \((x_2, y_2) = (1, 17)\). Substituting the values: \[ \frac{y + 13}{17 + 13} = \frac{x + 8}{1 + 8} \] This simplifies to: \[ \frac{y + 13}{30} = \frac{x + 8}{9} \] Cross-multiplying gives: \[ 9(y + 13) = 30(x + 8) \] Expanding both sides: \[ 9y + 117 = 30x + 240 \] Rearranging the equation: \[ 30x - 9y + 123 = 0 \quad \text{(Equation 2)} \] ### Step 3: Analyze the equations Now we have two equations: 1. \(10x + 3y - 1 = 0\) 2. \(30x - 9y + 123 = 0\) To check if these lines are the same, we can manipulate Equation 1: Multiplying Equation 1 by 3: \[ 30x + 9y - 3 = 0 \] Now, if we compare this with Equation 2: \[ 30x - 9y + 123 = 0 \] We can see that: - The coefficients of \(x\) are the same. - The coefficients of \(y\) are opposite. - The constant terms differ. This indicates that the two lines are not the same but are parallel. ### Step 4: Conclusion Since the two lines are parallel, they do not intersect at any point. Therefore, the answer to the question is that the two line segments do not cut each other at any point.
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A DAS GUPTA-Coordinates and Straight Lines-EXERCISE
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  2. If the sum of the distances of a point from two perpendicular lines in...

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  3. The image of the point (-1,3) by the line x - y = 0 , is

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  4. The point (4, 1) undergoes the following three transformations success...

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  5. All points lying inside the triangle formed by the points (1,3) ,(5,0)...

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  6. The two line segment joining (-2,7), (-5,-3) and (-8,-13),(1,17) cut e...

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  7. If line y - x + 2 = 0 is shifted parallel to itself towards the x-axis...

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  8. If the point (a^2,a+1) lies in the angle between the lines 3x-y+1=0 an...

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  9. The locus of a point which moves so that the difference of the squares...

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  10. If |x1y1 1x2y2 1x3y3 1|=|a1b1 1a2b2 1a3b3 1| then the two triangles ...

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  11. The straight line passing through the point of intersection of the str...

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  12. State whether the statements are true or false. The perpendicular bise...

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  13. P(m, n) (where m, n are natural numbers) is any point in the interior ...

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  14. If alpha,alpha^2) falls inside the angle made by the lines 2y=x ,x >0&...

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  15. The image of the point A (1,2) by the line mirror y=x is the point B a...

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  16. A=(sqrt(1-t^2)+t, 0) and B=(sqrt(1-t^2)-t, 2t) are two variable points...

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  17. If one diagonal of a square is the portion of the line x/a + y/b = 1 ...

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  18. The three intercepts made on the line x+y=5sqrt(2) by the lines y=x ta...

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  19. Let the coordinates of the foot of the perpendicular from the vertices...

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  20. P is any point on the x-a=0. If A=(a,0)and PQ , the bisector of angleO...

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