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Line Equation Examples

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 = (x₁ + x₂)/2 and y = (y₁ + y₂)/2; So, the coordinates of the midpoint are; x = (x₁ + x₂)/2 = [3 + (-1)]/2 = (3 - 1)/2 = 2/2 = 1; y = (y₁ + y₂)/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 L₁ which passing through point ( -3, -1 ) and is parallel to L₂ , whose equatoion is 2x - y - 3 = 0.

Solution
Step 1:: Obtain the slope k₂ of line L₂ 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 L₂ is k₂ = 2.
Step 2:: Two different nonvertical lines with slope k₁ and k₂ are parallel if and only if k₁ = k₂ .; Thus the slope of line L₁ is k₁ = 2.
Step 3:: To obtain line equation L₁ in slope-intercept form, using the formula:; y - y₀ = k(x - x₀).; Given x₀ = -3 and y₀ = -1,; then y - (-1) = 2[x - (-3)]; y + 1 = 2(x + 3); y + 1 = 2x + 6; y = 2x + 5.; The general form of line L₁ is: 2x - y + 5 = 0.

Question 4: Line L₁ contains the point ( -3, 2 ) and is perpendicular to the L₂, whose equatoion is 3x + 2y - 6 = 0.

Solution
Step 1:: Obtain the slope k₂ of L₂ 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 L₂ is k₂ = -3/2.
Step 2:: Two nonvertical lines with slope k₁ and k₂ are perpendicular if and only if k₁ k₂ = -1,; or, equivalently, k₁ = - 1/k₂.; Thus the slope of L₁ is k₁ = -1/(-3/2) = 2/3.
Step 3:: To obtain line equation L₁ in slope-intercept form, using the formula:; y - y₀ = k(x - x₀).; Given x₀ = -3 and y₀ = 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 L₁ is: 2x - 3y + 12 = 0.