Question

find the value of k for which the points with coordinates (2, 5),(k,11/8)and (4,6) are collinear​

Answer 1

Step-by-step explanation:

Since given points are collinear,

Therefore their SLOPES will be equal.

Hence,

$\frac{5 - \frac{11}{8} }{2 - k} = \frac{ \frac{11}{8} - 6 }{k - 4} \\ \implies \: \frac{\frac{5 \times 8 - 11}{8} }{2 - k} = \frac{ \frac{11 - 8 \times 6}{8}}{k - 4} \\ \implies \: \frac{\frac{40 - 11}{8} }{2 - k} = \frac{ \frac{11 - 48}{8}}{k - 4} \\ \implies \: \frac{\frac{29}{8} }{2 - k} =\implies \frac{ \frac{ - 37}{8}}{k - 4} \\ \frac{29}{8(2 - k)} = - \frac{37}{8(k - 4)} \\ \implies\frac{29}{2 - k} = - \frac{37}{k - 4} \: \\ \implies \: 29(k - 4) = - 37(2 - k) \\ \implies \: 29k - 116 = - 74 + 37 k \\ \implies \: 29k - 116 = - 74 + 37 k \\ \implies \: 74 - 116 = 37 k -29k \\ \implies- 42 = 8 k \\ \implies \: k = \frac{ - 42}{8} \\ \implies \huge \fbox {k = \frac{ -21}{4} }$