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प्रश्न
If the points A(k + 1, 2k), B(3k, 2k + 3) and C(5k − 1, 5k) are collinear, then find the value of k
उत्तर
Since the given points are collinear, the area of the triangle formed by them must be 0.
`=> 1/2 |x_1(y_2-y_3) + x_2(y_3 - y_1) +x_3(y_1 - y_2)| = 0`
Here
x1=k + 1, y1 = 2k;
x2 = 3k, y2 = 2k + 3 and
x3 = 5k − 1, y3=5k
∴ `1/2|(k+1){(2k+3)−(5k)}+(3k){(5k)−(2k)}+(5k−1){(2k)−(2k+3)}|=0`
⇒ (k+1){−3k+3}+(3k){3k}+(5k−1){−3}=0
⇒ −3k2 + 3k − 3k + 3 + 9k2 − 15k + 3=0
⇒ 6k2 − 15k + 6 = 0
⇒ 2k2 − 5k + 2 = 0
⇒ 2k2 − 4k − k + 2 = 0
⇒ 2k(k − 2) −1(k − 2) = 0
⇒(k − 2)(2k − 1) = 0
`=> k = 1/2, 2`
So the value of k are `1/2` and 2
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