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Line Equation Examples

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

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 square root of (1 + 1) = square root of 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:
How to solve two variables equation? What is the distance from the origin to the point.
Therefore, the distance to the origin is 1.414 units.

What is the Midpoint of a segment?

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)
Find the midpoint of line segment

What is the line equation when two lines parallel?

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.

What is the line equation when two lines are perpendicular?

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.