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# Line Equation Examples

### What is the distance between the crossing point of two lines and the origin?

### What is the Midpoint of a segment?

### What is the line equation when two lines parallel?

### What is the line equation when two lines are perpendicular?

- Question 1
- The crossing point between the two lines, 2x - y - 1 = 0 and x + 2y - 3 = 0 is d distance away from the origin. What is the value of d?

- Solution
- Find the crossing point by treating these two lines with regard to their corresponding equations. Multiplying the first equation by 2 and adding it to the second will give you 5x = 5, or x = 1. Then, plugging this value into either equation will give y = 1, so the point of crossing is (1, 1) and its distance to the origin is Sqrt (1 + 1) = Sqrt (2) = 1.414
- 2x - y - 1 = 0 ...equation1
- x + 2y - 3 = 0 ...equation2
- equation1 × 2: 4x - 2y - 2 = 0 ...equation3
- equation3 + equation2: 4x + x
~~- 2y~~+~~2y~~- 2 + (-3) = 0 - 5x - 5 = 0
- 5x = 5
- x = 1
- from equation1: y = 2x - 1 = 2 × 1 - 1 = 1
- the solutions are: x = 1 and y = 1
- The distance from (x, y) = (1, 1) to the origin is:
- Therefore, the distance to the origin is 1.414 units.

- Question 2
- Where are the coordinates of the midpoint M of the segment joing the pair of points (3, 5) and (-1, -7)?

- Solution
- The midpoint M(x, y) of the line segment is given by x = (x
_{1}+ x_{2})/2 and y = (y_{1}+ y_{2})/2 - So, the coordinates of the midpoint are
- x = (x
_{1}+ x_{2})/2 = [3 + (-1)]/2 = (3 - 1)/2 = 2/2 = 1 - y = (y
_{1}+ y_{2})/2 = [5 + (-7)]/2 = (5 - 7)/2 = -2/2 = -1 - Therefore, the coordinate of the midpoint is (1, -1)

- Question 3
- Obtain the equation of the line L
_{1}which passing through point ( -3, -1 ) and is parallel to L_{2}, whose equatoion is 2x - y - 3 = 0.

- Solution
- Step 1:
- Obtain the slope k
_{2}of line L_{2}by solving the equation 2x - y - 3 = 0 for y in terms of x. - y = 2x - 3.
- The slope-intercept form for one line is:
- y = kx + b, then the slope of L
_{2}is k_{2}= 2. - Step 2:
- Two different nonvertical lines with slope k
_{1}and k_{2}are parallel if and only if k_{1}= k_{2}. - Thus the slope of line L
_{1}is k_{1}= 2. - Step 3:
- To obtain line equation L
_{1}in slope-intercept form, using the formula: - y - y
_{0}= k(x - x_{0}). - Given x
_{0}= -3 and y_{0}= -1, - then y - (-1) = 2[x - (-3)]
- y + 1 = 2(x + 3)
- y + 1 = 2x + 6
- y = 2x + 5.
- The general form of line L
_{1}is: 2x - y + 5 = 0.

- Question 4
- Line L
_{1}contains the point ( -3, 2 ) and is perpendicular to the L_{2}, whose equatoion is 3x + 2y - 6 = 0.

- Solution
- Step 1:
- Obtain the slope k
_{2}of L_{2}by solving the equation 3x + 2y - 6 = 0 for y in terms of x. - 2y = -3x + 6
- y = -(3/2)x + 3
- The slope-intercept form for one line is:
- y = kx + b, then the slope of L
_{2}is k_{2}= -3/2. - Step 2:
- Two nonvertical lines with slope k
_{1}and k_{2}are perpendicular if and only if k_{1}k_{2}= -1, - or, equivalently, k
_{1}= - 1/k_{2}. - Thus the slope of L
_{1}is k_{1}= -1/(-3/2) = 2/3. - Step 3:
- To obtain line equation L
_{1}in slope-intercept form, using the formula: - y - y
_{0}= k(x - x_{0}). - Given x
_{0}= -3 and y_{0}= 2, - then y - 2 = 2/3[x - (-3)]
- y - 2 = 2/3(x + 3)
- y - 2 = (2/3)x + 2
- y = (2/3)x + 4
- both sides multiply 3 to remove denominator
- 3y = 2x + 12.
- The general form of line L
_{1}is: 2x - 3y + 12 = 0.

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