Show that the line joining A(−1, 1) and B(−9, 6) are parallel to the line joining P(−2, 14) and Q(6, 9)

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Solution

We know that slope of a line = $\frac{y_{2} - y_{1}}{x_{2} - x_{1}} = \frac{y_{1} - y_{2}}{x_{1} - x_{2}}$

Slope of the line joining the points A(−1, 1) and B(−9, 6) = $\frac{1 - 6}{- 1 - (- 9)} = - \frac{5}{8}$

Slope of the line joining the points P(−2, 14) and Q(6, 9) = $\frac{14 - 9}{- 2 - 6} = - \frac{5}{8}$

Here, slope of line AB = slope of line PQ
∴ Line AB is parallel to PQ.

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Show that the line joining A(−1, 1) and B(−9, 6) are parallel to the line joining P(−2, 14) and Q(6, 9)

We know that slope of a line = y2-y1x2-x1=y1-y2x1-x2 Slope of the line joining the points A(−1, 1) and B(−9, 6) = 1-6-1-(-9)=-58 Slope of the line joining the points P(−2, 14) and Q(6, 9) = 14-9-2-6=-58 Here, slope of line AB = slope of line PQ ∴ Line AB is parallel to PQ.