Find k, if PQ || RS and P(2, 4), Q (3, 6), R(3, 1), S(5, k) .
Find k, if PQ || RS and P(2, 4), Q (3, 6), R(3, 1), S(5, k) .
Given P(2, 4), Q (3, 6), R(3, 1), S(5, k)
PQ || RS so, slope of PQ = slope of RS
\[\frac{6 - 4}{3 - 2} = \frac{k - 1}{5 - 3}\]
\[ \Rightarrow \frac{2}{1} = \frac{k - 1}{2}\]
\[ \Rightarrow k - 1 = 4\]
\[ \Rightarrow k = 5\]
Explanation:-
The problem gives us four points in a 2-dimensional space, namely P(2, 4), Q(3, 6), R(3, 1), and S(5, k). The problem tells us that PQ is parallel to RS and asks us to find the value of k.
Two lines are parallel if and only if their slopes are equal. Therefore, if PQ is parallel to RS, then the slope of PQ is equal to the slope of RS. We can use the formula for slope to find the slope of PQ and RS:
slope of PQ = (y-coordinate of Q – y-coordinate of P) / (x-coordinate of Q – x-coordinate of P)
slope of RS = (y-coordinate of S – y-coordinate of R) / (x-coordinate of S – x-coordinate of R)
Since PQ is parallel to RS, their slopes must be equal. Therefore, we can set the two formulas for slope equal to each other and solve for k:
(y-coordinate of Q – y-coordinate of P) / (x-coordinate of Q – x-coordinate of P) = (y-coordinate of S – y-coordinate of R) / (x-coordinate of S – x-coordinate of R)
(6 – 4) / (3 – 2) = (k – 1) / (5 – 3)
2 = (k – 1) / 2
4 = k – 1
k = 5
Therefore, the value of k is 5.
Chapter 5. Co-ordinate Geometry – Practice Set 5.3 (Page 122)