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Algebra
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:
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 = (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)
What is the line equation when two lines parallel?
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.
What is the line equation when two lines are perpendicular?
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.