Question

The distance between the Points (a SinA, a CosA) and (a CosA, -a SinA ) is

Answer 1

$\textsf{\large{\underline{Solution}:}}$

Let P(x₁, y₁) and Q(x₂, y₂) be two points in the coordinate plane. Then, the distance between P and Q is calculated as:

$\rm: \longmapsto D = \sqrt{ {(x_{1} - x_{2})}^{2} +{(y_{1} - y_{2})}^{2} }$

Here:

$\rm: \longmapsto x_{1} = a \sin A$

$\rm: \longmapsto x_{2} = a \cos A$

$\rm: \longmapsto y_{1} = a \cos A$

$\rm: \longmapsto y_{2} = - a \sin A$

Therefore, the distance between the two points is given as:

$\rm: \longmapsto D = \sqrt{ {(a \sin A- a \cos A)}^{2} +{(a \cos A + a \sin A)}^{2} }$

We know that:

$\rm: \longmapsto {(x + y)}^{2} + {(x - y)}^{2} = 2( {x}^{2} + {y}^{2} )$

Therefore:

$\rm: \longmapsto D = \sqrt{2( {a}^{2} \sin^{2}A + {a}^{2} \cos^{2} A)}$

$\rm: \longmapsto D = \sqrt{2{a}^{2}( \sin^{2}A + \cos^{2} A)}$

We know that:

$\rm: \longmapsto\sin^{2}A + \cos^{2} A = 1$

Therefore:

$\rm: \longmapsto D = \sqrt{2{a}^{2} \times 1}$

$\rm: \longmapsto D = \sqrt{2{a}^{2}}$

$\rm: \longmapsto D =a \sqrt{2} \: \: units.$

★ Which is our required answer.

$\textsf{\large{\underline{More To Know}:}}$

1. Section formula.

Let P(x₁, y₁) and Q(x₂, y₂) be two points in the coordinate plane and R(x, y) be the point which divides PQ internally in the ratio m₁ : m₂. Then, the coordinates of R will be:

$\rm:\longmapsto R = \bigg(\dfrac{m_{1}x_{2}+m_{2}x_{1}}{m_{1}+m_{2}}, \dfrac{m_{1}y_{2}+m_{2}y_{1}}{m_{1}+m_{2}}\bigg)$

2. Mid-point formula.

Let P(x₁, y₁) and Q(x₂, y₂) be two points in the coordinate plane and R(x, y) be the mid-point of PQ. Then, the coordinates of R will be:

$\rm:\longmapsto R =\bigg(\dfrac{x_{1}+x_{2}}{2}, \dfrac{y_{1}+y_{2}}{2}\bigg)$

3. Centroid of a triangle.

Centroid of a triangle is the point where the medians of the triangle meet.

Let A(x₁, y₁), B(x₂, y₂) and C(x₃, y₃) be the vertices of a triangle. Let R(x, y) be the centroid of the triangle. Then, the coordinates of R will be:

$\rm:\longmapsto R= \bigg(\dfrac{x_{1}+x_{2}+x_{3}}{3}, \dfrac{y_{1}+y_{2}+y_{3}}{3}\bigg)$