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 (5, -1), Q (6, 11), R (6, -4k) and S (7, $k^{2}$ )

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Solution

Since, PQ || RS,

Slope of PQ = Slope of RS

(i) Slope of PQ = $fraction numerator 6 minus 4 over denominator 3 minus 2 end fraction equals 2$

Slope of RS = $fraction numerator k minus 1 over denominator 10 minus 8 end fraction equals fraction numerator k minus 1 over denominator 2 end fraction$

$2 equals fraction numerator k minus 1 over denominator 2 end fraction k minus 1 equals 4 k equals 5$

(ii) Slope of PQ = $fraction numerator 11 plus 1 over denominator 6 minus 5 end fraction equals 12$

Slope of RS = $fraction numerator k squared plus 4 k over denominator 7 minus 6 end fraction equals k squared plus 4 k$

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Similar questions

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, $k^{2}$ )

k

P (2, 4), Q (3, 6), R (8, 1)

S (10, k)

P (2, 4) Q (3, 6), R (3, 1), S (5, k)

Find the value of k if PQ and RS are parallel, given P(1, 2), Q(4,5) and R(3, 4), S(8, k).

n \sum x = 1

p n^{4} + q n^{3} + r n^{2} + \frac{7}{3} n + s

i) p = \frac{1}{2}, q = \frac{8}{3} i i) p = \frac{1}{3}, q = \frac{1}{4} i i i) r = \frac{9}{2}, s = 0 i v) r = \frac{5}{6}, s = 0

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