Determine if the points (1, 5), (2, 3) and (– 2, – 11) are collinear
Let the points (1, 5), (2, 3), and (−2, −11) be representing the vertices A, B, and C of the given triangle respectively.
Let, A=(1,5); B=(2, 3); C=(−2, −11)
\(\therefore\) AB=\(\sqrt{(1-2)^2+(5-3)^2}=\sqrt5\)
BC=\(\sqrt{(2-(-2))^2+(3-(-11))^2}=\sqrt{4^2+14^2}=\sqrt{16+196}=\sqrt{212}\)
CA=\(\sqrt{(1-(-2))^2+(5-(-11))^2}=\sqrt{3^2+16^2}=\sqrt{9+256}=\sqrt{265}\)
Since, AB+BC\(\neq\)CA
Therefore, the points (1, 5), (2, 3), and (−2, −11) are not collinear.
In a classroom, 4 friends are seated at points A, B, C and D. Champa and Chameli walk into the class and after observing for a few minutes Champa asks Chameli, “Don’t you think ABCD is a square?” Chameli disagrees. Using the distance formula, find which of them is correct.
Find the point on the x-axis which is equidistant from (2, –5) and (–2, 9).
| Case No. | Lens | Focal Length | Object Distance |
|---|---|---|---|
| 1 | \(A\) | 50 cm | 25 cm |
| 2 | B | 20 cm | 60 cm |
| 3 | C | 15 cm | 30 cm |