Question

Find the value of ‘K’ for which the points are collinear.(i) (7, −2) (5, 1) (3, K) (ii) (8, 1), (K, −4), (2, −5)(iii) (K, K) (2, 3) and (4, −1).

Answer 1

if three points $\bf{(x_1,y_1),(x_2,y_2)\:and\:(x_3,y_3)}$ are collinear.
then, $\bf{x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)= 0}$

(i) (7, -2) , (5, 1) and (3, K)
=> 7(1 - k) + 5(k +2) + 3(-2 - 1) = 0
=> 7 - 7k + 5k + 10 - 9 = 0
=> -2k + 8 = 0
=> k = 4

(ii) (8, 1) , (k,-4) and (2,-5)
=> 8(-4 + 5) + k(-5 - 1) + 2(1 + 4) = 0
=> 8 × 1 -6k + 10 = 0
=> 8 - 6k + 10 = 0.
=> k = 3

(iii) (k,k), (2,3) and (4,-1)
=> k(3 + 1) + 2(-1 - k) + 4(k - 3) = 0
=> 4k -2 - 2k + 4k - 12 = 0
=> 6k - 14 = 0
=> k = 7/3

Answer 2

Answer:

Step-by-step explanation:

From a point Q, the length of the tangent to a circle is 24 cm. and the distance of Q from the centre is 25 cm. The radius of the circle is

(a) 7 cm

(b) 12 cm

(c) 15 cm

(d) 24.5cm