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Find the value(s) of k so that PQ will be parallel to RS. Given :

(i) P(2, 4), Q(3, 6), R(8, 1) and S(10, k)

(ii) P(3, -1), Q(7, 11), R(-1, -1) and S(1, k)

(iii) P(5, -1), Q(6, 11), R(6, -4k) and S(7, k2)

Straight Line Eq

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Answer

Since, PQ is parallel to RS. So, slope of PQ will be equal to slope of RS.

(i) P(2, 4), Q(3, 6), R(8, 1) and S(10, k)

Slope of PQ = Slope of RS

∴6−43−2=k−110−8⇒21=k−12⇒4=k−1⇒k=4+1=5.\therefore \dfrac{6 - 4}{3 - 2} = \dfrac{k - 1}{10 - 8} \\[1em] \Rightarrow \dfrac{2}{1} = \dfrac{k - 1}{2} \\[1em] \Rightarrow 4 = k - 1 \\[1em] \Rightarrow k = 4 + 1 = 5.

Hence, k = 5.

(ii) P(3, -1), Q(7, 11), R(-1, -1) and S(1, k)

Slope of PQ = Slope of RS

∴11−(−1)7−3=k−(−1)1−(−1)⇒124=k+12⇒3=k+12⇒k+1=6⇒k=5.\therefore \dfrac{11 - (-1)}{7 - 3} = \dfrac{k - (-1)}{1 - (-1)} \\[1em] \Rightarrow \dfrac{12}{4} = \dfrac{k + 1}{2} \\[1em] \Rightarrow 3 = \dfrac{k + 1}{2} \\[1em] \Rightarrow k + 1 = 6 \\[1em] \Rightarrow k = 5.

Hence, k = 5.

(iii) P(5, -1), Q(6, 11), R(6, -4k) and S(7, k2)

Slope of PQ = Slope of RS

∴11−(−1)6−5=k2−(−4k)7−6⇒121=k2+4k1⇒12=k2+4k⇒k2+4k−12=0⇒k2+6k−2k−12=0⇒k(k+6)−2(k+6)=0⇒(k−2)(k+6)=0⇒k−2=0 or k+6=0⇒k=2 or k=−6.\therefore \dfrac{11 - (-1)}{6 - 5} = \dfrac{k^2 - (-4k)}{7 - 6} \\[1em] \Rightarrow \dfrac{12}{1} = \dfrac{k^2 + 4k}{1} \\[1em] \Rightarrow 12 = k^2 + 4k \\[1em] \Rightarrow k^2 + 4k - 12 = 0 \\[1em] \Rightarrow k^2 + 6k - 2k - 12 = 0 \\[1em] \Rightarrow k(k + 6) - 2(k + 6) = 0 \\[1em] \Rightarrow (k - 2)(k + 6) = 0 \\[1em] \Rightarrow k - 2 = 0 \text{ or } k + 6 = 0 \\[1em] \Rightarrow k = 2 \text{ or } k = -6.

Hence, k = 2 or -6.

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