Mathematics

Straight Lines

Posted by Asok K Nadhani on

Last Updated on: 16th February 2024, 11:03 am

Straight Lines

Straight Lines – Important Formula

  • Distance between the two points P (x1,y1) and Q (x2,y2) : PQ=√[(x2-x1)2+(y2-y1)2]
  • Co-ordinates of a point internally dividing the line segment joining the two points (x1,y1) and (x2,y2), in the ratio m:n is  (my2+ny1/(m+n). So, when m=n, the co-ordinate of mid-point of the segment is (x1+x2)/2, (y1+y2)/2
  • Area of triangle PQR with vertices P (x1,y1), Q (x2,y2), R (x3,y3) is |x1(y2-y3) + (x2(y3-y1) +x3(y1-y2)| When the 3 points are collinear, the Area of Triangle is 0

Slope of a Line

A line in a coordinate plane forms two supplementary angles (\displaystyle \theta ) and (180 – \displaystyle \theta ) with the x-axis.

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F-1:Slope of Line

  • Inclination of Line : The angle q made by the line l with positive direction of x-axis and measured anti clockwise is called the inclination of the line (F-1). So, 0°≤𝜃 ≤180°. Lines parallel to x-axis, or coinciding with x-axis, have inclination of 0°. The inclination of a vertical line (parallel to or coinciding with y-axis) is 90°.
  • Gradient (or slope) of Line : The Gradient of Line with inclination of angle 𝜃 is tan𝜃 (denoted by m). So, m= tan𝜃 (𝜃≠ 90°) Slope of line with inclination 90  is undefined. So, slope of x axis is 0 and slope of y axis is undefined.

Slope of line passing through 2 given points

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F-2A:Slope of line passing through 2 given points, Acute Inclination

Screenshot 1

F-2B: Slope of line passing through 2 given points, Obtuse through 2 given points, Acute Inclination

P(x1,y1) and Q(x2, y2) are two points on non-vertical line l whose inclination is 𝜃 (F-2). So, x1≠x2, (otherwise the line will become perpendicular to x-axis and its slope would be undefined). The inclination of the line l may be acute or obtuse. We would take up take both cases.

Draw perpendicular QR to x-axis and PM perpendicular to RQ.

  • Inclination 𝜃 Acute : \displaystyle \angle MPQ = 𝜃 … (1). So, slope of line l = m = tan 𝜃. In \displaystyle \vartriangle MPQ, \displaystyle \frac{{MQ}}{{MP}}= y2-y1/x2-x1 …. (2)
  • From the eq (1) & (2), we get, m=y2-y1/x2-x1
  • Inclination 𝜃 Obtuse : \displaystyle \angle MPQ = 1800 – 𝜃, So, 𝜃=1800 – \displaystyle \angle MPQ. So, slope of line l, m=tan(1800 – \displaystyle \angle MPQ) = -tan \displaystyle \angle MPQ =\displaystyle -\frac{{MQ}}{{MP}} = – (y2-y1)/(x2-x1) =y2-y1/x2-x1 .

So, in both cases, slope of the line through the points (x1,y1) and (x2, y2)=