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A line through origin meets the line x =...

A line through origin meets the line `x = 3y + 2` at right angles at point X. Find the co-ordinates of X. 

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To solve the problem step by step, we will find the coordinates of the point X where the line through the origin meets the line \( x = 3y + 2 \) at right angles. ### Step 1: Identify the equations of the lines The line through the origin can be represented as: \[ y = mx \] where \( m \) is the slope of the line. The given line is: \[ x = 3y + 2 \] ### Step 2: Convert the given line to slope-intercept form To find the slope of the given line, we need to rearrange it into the form \( y = mx + c \). Starting from: \[ x = 3y + 2 \] we can rearrange it as follows: \[ 3y = x - 2 \] \[ y = \frac{1}{3}x - \frac{2}{3} \] From this, we can see that the slope \( m_1 \) of the line \( x = 3y + 2 \) is: \[ m_1 = \frac{1}{3} \] ### Step 3: Determine the slope of the line through the origin Since the two lines are perpendicular, the product of their slopes must equal -1. Let the slope of the line through the origin be \( m \): \[ m \cdot m_1 = -1 \] \[ m \cdot \frac{1}{3} = -1 \] Solving for \( m \): \[ m = -3 \] ### Step 4: Write the equation of the line through the origin Now that we have the slope, the equation of the line through the origin is: \[ y = -3x \] ### Step 5: Solve the two equations simultaneously We need to find the point of intersection of the lines \( y = -3x \) and \( y = \frac{1}{3}x - \frac{2}{3} \). We can set them equal to each other: \[ -3x = \frac{1}{3}x - \frac{2}{3} \] ### Step 6: Clear the fractions To eliminate the fraction, multiply the entire equation by 3: \[ -9x = x - 2 \] ### Step 7: Rearrange the equation Rearranging gives: \[ -9x - x = -2 \] \[ -10x = -2 \] Dividing by -10: \[ x = \frac{1}{5} \] ### Step 8: Find the y-coordinate Now substitute \( x = \frac{1}{5} \) back into the equation of the line through the origin to find \( y \): \[ y = -3\left(\frac{1}{5}\right) = -\frac{3}{5} \] ### Step 9: Write the coordinates of point X Thus, the coordinates of point X where the two lines meet at right angles are: \[ \left( \frac{1}{5}, -\frac{3}{5} \right) \] ### Summary of the solution The coordinates of point X are: \[ \left( \frac{1}{5}, -\frac{3}{5} \right) \]
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ICSE-EQUATION OF A LINE-EXERCISE 14(E)
  1. Show that A (3, 2), B (6, -2) and C (2, -5) can be the vertices of a s...

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  2. Show that A (3, 2), B (6, -2) and C (2, -5) can be the vertices of a s...

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  3. A line through origin meets the line x = 3y + 2 at right angles at poi...

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  4. A straight line passes through the point (3, 2) and the portion of thi...

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  5. Find the equation of the line passing through the point of intersectio...

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  6. Find the equation of the line which is perpendicular to the line x/a -...

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  7. O (0, 0), A (3, 5) and B (-5, -3) are the vertices of triangle OAB. Fi...

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  8. O (0, 0), A (3, 5) and B (-5, -3) are the vertices of triangle OAB. Fi...

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  9. Determine whether the line through points (-2, 3) and (4, 1) is perpen...

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  10. Given a straight line x cos 30^@ + y sin 30^@ = 2. Determine the equat...

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  11. Find the value of k such that the line (k-2)x+(k+3)y-5=0 perpendi...

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  12. Find the value of k such that the line (k-2)x+(k+3)y-5=0 is para...

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  13. The vertices of a triangle are A (0, 5), B (-1, -2) and C (11, 7). Wri...

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  14. The vertices of a triangle are A (0, 5), B (-1, -2) and C (11, 7). Wri...

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  15. From the given figure, find : the co ordinates of A, B and C.

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  16. From the given figure, find : the equation of the line through A ...

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  17. P(3, 4), Q(7, -2) and R(-2, -1) are the vertices of triangle PQR. Writ...

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  18. A(8, -6), B(-4, 2) and C(0, -10) are vertices of a triangle ABC. If P ...

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  19. In the given figure, line APB meets the x-axis at point A and y-axis a...

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  20. A line AB meets the x-axis at point A and y-axis at point B. The point...

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