We will discuss here about the slope of the line joining two points.
To find the slope of a non-vertical straight line passing through two given fixed points:
Let P (x\(_{1}\), y\(_{1}\)) and Q (x\(_{2}\), y\(_{2}\)) be the two given points. According to the problem, the straight line PQ is non-vertical x\(_{2}\) ≠ x\(_{1}\).
Required to find, the slope of the line through P and Q.
From P, Q draw perpendiculars PM, QN on x-axis and PL ⊥ NQ. Let θ be the inclination of the line PQ, then ∠LPQ = θ.
From the above diagram, we have
PL = MN = ON - OM = x\(_{2}\) - x\(_{1}\) and
LQ = = NQ - NL = NQ - MP = y\(_{2}\) - y\(_{1}\)
Therefore, the slope of the line PQ = tan θ
= \(\frac{LQ}{PL}\)
= \(\frac{y_{2} - y_{1}}{x_{2} - x_{1}}\)
= \(\frac{Difference\, of\, ordinates\,of\, the\, given\, points}{Difference\, of\, their\, abscissae}\)
Hence, the slope (m) of a non-vertical line passing through the points P (x\(_{1}\), y\(_{1}\)) and Q (x\(_{2}\), y\(_{2}\)) is given by
slope = m = \(\frac{y_{2} - y_{1}}{x_{2} - x_{1}}\)
1. Find the slope of the line passing through the points M (-2, 3) and N (2, 7).
Solution:
Let M (-2, 3) = (x\(_{1}\), y\(_{1}\)) and N (2, 7) = (x\(_{2}\), y\(_{2}\))
We know that the slope of a straight line passing through two points (x\(_{1}\), y\(_{1}\)) and (x\(_{2}\), y\(_{2}\)) is
m = \(\frac{y_{2} - y_{1}}{x_{2} - x_{1}}\)
Therefore, slope of MN = \(\frac{y_{2} - y_{1}}{x_{2} - x_{1}}\) = \(\frac{7 - 3}{2 + 2}\) = \(\frac{4}{4}\) = 1.
2. Find the slope of the line passing through the pairs of points (-4, 0) and origin.
Solution:
We know that the coordinate of the origin is (0, 0)
Let P (-4, 0) = (x\(_{1}\), y\(_{1}\)) and O (0, 0) = (x\(_{2}\), y\(_{2}\))
We know that the slope of a straight line passing through two points (x\(_{1}\), y\(_{1}\)) and (x\(_{2}\), y\(_{2}\)) is
m = \(\frac{y_{2} - y_{1}}{x_{2} - x_{1}}\)
Therefore, slope of PO = \(\frac{y_{2} - y_{1}}{x_{2} - x_{1}}\)
= \(\frac{0 - (0}{0 - (- 4)}\)
= \(\frac{0}{4}\)
= 0.
● Equation of a Straight Line
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