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If the points (-2,6) and (4,8) is perpen...

If the points (-2,6) and (4,8) is perpendicular to the line joining the points (8,12) and (x,24) then the value of x is

A

(a) `2/7`

B

(b) `4`

C

(c) `3`

D

(d) `-4`

Text Solution

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The correct Answer is:
To solve the problem, we need to find the value of \( x \) such that the line joining the points (-2, 6) and (4, 8) is perpendicular to the line joining the points (8, 12) and (x, 24). ### Step-by-Step Solution: 1. **Identify the Points**: Let the points be defined as follows: - Point A: \( (-2, 6) \) (denote as \( (x_1, y_1) \)) - Point B: \( (4, 8) \) (denote as \( (x_2, y_2) \)) - Point C: \( (8, 12) \) (denote as \( (x_3, y_3) \)) - Point D: \( (x, 24) \) (denote as \( (x_4, y_4) \)) 2. **Calculate the Slope of Line AB**: The slope \( m_1 \) of line AB can be calculated using the formula: \[ m_1 = \frac{y_2 - y_1}{x_2 - x_1} = \frac{8 - 6}{4 - (-2)} = \frac{2}{6} = \frac{1}{3} \] 3. **Calculate the Slope of Line CD**: The slope \( m_2 \) of line CD can be calculated using the formula: \[ m_2 = \frac{y_4 - y_3}{x_4 - x_3} = \frac{24 - 12}{x - 8} = \frac{12}{x - 8} \] 4. **Set Up the Perpendicular Condition**: Since the lines are perpendicular, the product of their slopes must equal -1: \[ m_1 \cdot m_2 = -1 \] Substituting the slopes we found: \[ \frac{1}{3} \cdot \frac{12}{x - 8} = -1 \] 5. **Solve for x**: Multiply both sides by \( 3(x - 8) \) to eliminate the fraction: \[ 12 = -3(x - 8) \] Expanding the right side: \[ 12 = -3x + 24 \] Rearranging gives: \[ -3x = 12 - 24 \] \[ -3x = -12 \] Dividing by -3: \[ x = 4 \] ### Conclusion: The value of \( x \) is \( 4 \).
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ICSE-STRAIGHT LINES -Multiple Choice Questions
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  9. The equation of the line which cuts of intercept 4 on x-ais and makes ...

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  10. If the straight line y=mx+c passes through the points (2,4) and (-3,6)...

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  11. A line passes through P(1,2) such that the portion of the line interce...

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  14. The angle between lines y=(2-sqrt(3))x+5 and y=(2+sqrt(3))x+7 is

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  16. If the image of the point (-3,k) in the line 2x+y-2=0 is the point (1,...

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