如何判断两条线是否平行？
几何是处理点、角、线、线段等的数学分支之一。它们帮助我们确定两个空间之间的空间关系。从古代到发展中国家现代化系统的发展，几何学的方法都很明显。现代化系统完全依赖于几何，因为它用于设计、建筑工程、建筑材料的选择等等。
Geometry is a branch of mathematics that deals with the study of shapes and their properties.
给定的文章是关于平行线的讨论，用示例表示平行线的属性。该文章还包括用给定线的方程确定平行线的方法。连同一些示例数值问题供参考。
什么是平行线？
平行线被定义为彼此距离相等并在无穷远点处相交的一对线。平行线可以在两个方向上无限延伸。我们可以简单地将平行线理解为沿着地平线移动的谎言，而不是在相同或相反的方向上相交。
笔记本线、铁轨、斑马线是我们可以在周围观察到的平行线的一些现实例子。
如何判断两条线是否平行？
回答：
当每条线的斜率相等时，确定两条线平行。如果发现两条线的斜率比较相等，则认为这些线是平行的。
为此，首先我们必须确定线的方程并推导出它们的斜率。如果发现它们的斜率相等，则证明这些线是平行的。
例如
让我们分别考虑两行方程 6x – 4y = 25 和 9x – 6y = 12。并且，让它们的斜率是 m1 和 m2。
这里，
6x – 4y = 25………….(I)
9x – 6y = 12…………..(ii)
现在，对于 m1 以 y = mx + b 的形式求解方程 (I)
=>6x – 4y = 25
=>4y = 6x – 25
两边除以4
=>4y/4 = 6x/4 – 25/4
=>y = 3/2 x – 25/4
=>m1 = 3/2
同样，对于 m2 以 y=mx+b 的形式求解方程 (ii)
=>9x – 6y = 12
=>6y = 9x – 12
两边除以6
=>6y/6 = 9/6x – 12/6
=>y = 3/2x – 2
=>m2 = 3/2
因此，两条给定的线是平行的，因为 m1 = m2。
示例问题
问题 1. 确定两条线 6x – 4y = 12 和 3x – 2y = 5 是否平行。
解决方案：
Let us consider two lines with equations 6x-4y=12 and 3x-2y=5 respectively. And, let their slopes be m1 and m2.
Here,
6x-4y=12………….(I)
3x-2y=5……….(ii)
Now, for m1 solving equation (I) in the form of y=mx+b
=>6x-4y=12
=>4y=6x-12
Dividing on both sides by 4
=>4y/4=6x/4 -12/4
=>y=3/2 x- 12/4
=>m1=3/2
Again, for m2 solving equation(ii) in the form of y=mx+b
=>-3x+2y=5
=>2y=3x+5
Dividing on both sides by 2
=>2y/2=3x/2+5/2
=>m2=3/2
Hence, the two given lines are parallel as m1=m2.
问题2.判断2/3x+y=5/3和2/3x+y=1这两条线是否平行。
解决方案：
Let us consider two lines with equations 2/3x+y=5/3 and 2/3x+y=1 respectively. And, let their slopes be m1 and m2.
2/3x+y=5/3…………(I)
2/3x+y=1 …………..(ii)
Now, for m1 comparing equation (I) with the form of y=mx+b
=>2/3x+y=5/3
=>y=-2/3x+5/3
=>m1=-2/3
Again, for m2 comparing equation (ii) with the form of y=mx+b
=>2/3x+y=1
=>y=-2/3x+1
=>m2=-2/3
Hence, the two given lines are parallel as m1=m2.
问题3.判断2x-y=-5和2x-y=1这两条线是否平行。
解决方案：
Let us consider two lines with equations 2x-y=-5 and 2x-y=1 respectively. And, let their slopes be m1 and m2.
2x-y=-5 …………(I)
2x-y=1…………..(ii)
Now, for m1 comparing equation (I) with the form of y=mx+b
=>2x-y=5
=>y=2x+5
=>m1=2
Again, for m2 comparing equation (ii) with the form of y=mx+b
=>2x-y=1
=>y=2x-1
=>m2=2
Hence, the two given lines are parallel as m1=m2.