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Answer:

[tex]\boxed{\begin{aligned}& \:\sf \:Distance \: of \: point \: (3, - 4) \: from \:origin \: is \: 5 \: units \: \\ \\& \:\sf \:(1, - 2) \: is \: nearest \: to \: y - axis\end{aligned}} \: \\ \\ [/tex]

Step-by-step explanation:

[tex]\large\underline{\sf{Solution-3}}[/tex]

We have to find the distance of the point (3, - 4) from the origin.

So, using Distance Formula, we get

[tex]\sf \: Distance = \sqrt{ {(3 - 0)}^{2} + {( - 4 - 0)}^{2} } \\[/tex]

[tex]\sf \: Distance = \sqrt{ {(3)}^{2} + {( - 4)}^{2} } \\[/tex]

[tex]\sf \: Distance = \sqrt{ 9 + 16} \\[/tex]

[tex]\sf \: Distance = \sqrt{ 25} \\[/tex]

[tex]\implies\sf \: Distance = 5 \: units \\ [/tex]

Hence,

[tex]\implies\sf \:\boxed{\sf \: Distance \: of \: point \: (3, - 4) \: from \:origin \: is \: 5 \: units \: } \\ \\ [/tex]

[tex]\large\underline{\sf{Solution-4}}[/tex]

We know, Distance of the point (a, b) from y - axis is a units

So,

[tex]\sf \: Distance\:of\:point\:(2,1)\:from\:y-axis = 2 \: unit\\ [/tex]

[tex]\sf \: Distance\:of\:point\:(1, - 2)\:from\:y-axis = 1 \: unit\\ [/tex]

[tex]\sf \: Distance\:of\:point\:(2, - 1)\:from\:y-axis = 2 \: unit \\ [/tex]

[tex]\sf \: Distance\:of\:point\:(2, - 2)\:from\:y-axis = 2 \: unit \\ [/tex]

Thus from above calculations, we concluded that

[tex]\implies\sf \: \boxed{\sf \: (1, - 2) \: is \: nearest \: to \: y - axis \: } \\ \\ [/tex]

[tex]\rule{190pt}{2pt}[/tex]

Formula Used:

Distance Formula :

Let A(x₁, y₁) and B(x₂, y₂) be two points in the cartesian plane, then distance between A and B is given by

[tex]\begin{gathered}\boxed{\tt{ AB \: = \sqrt{ {(x_{2} - x_{1}) }^{2} + {(y_{2} - y_{1})}^{2} }}} \\ \end{gathered} \\ [/tex]

[tex]\rule{190pt}{2pt}[/tex]

[tex] { \large{ \mathfrak{Additional\:Information}}}[/tex]

1. Section formula

Let A(x₁, y₁) and B(x₂, y₂) be two points in the cartesian plane and C(x, y) be the point which divides AB internally in the ratio m₁ : m₂, then the coordinates of C is given by

[tex]\begin{gathered} \boxed{\tt{ (x, y) = \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)}} \\ \end{gathered} \\ [/tex]

2. Mid-point formula

Let A(x₁, y₁) and B(x₂, y₂) be two points in the coordinate plane and C(x, y) be the mid-point of AB, then the coordinates of C is given by

[tex]\begin{gathered}\boxed{\tt{ (x,y) = \bigg(\dfrac{x_{1}+x_{2}}{2}, \dfrac{y_{1}+y_{2}}{2}\bigg)}} \\ \end{gathered} \\ [/tex]

3. Centroid of a triangle

Centroid of a triangle is defined as the point at which the medians of the triangle meet and is represented by the symbol G.

Let A(x₁, y₁), B(x₂, y₂) and C(x₃, y₃) be the vertices of a triangle and G(x, y) be the centroid of the triangle, then the coordinates of G is given by

[tex]\begin{gathered}\boxed{\tt{ (x, y) = \bigg(\dfrac{x_{1}+x_{2}+x_{3}}{3}, \dfrac{y_{1}+y_{2}+y_{3}}{3}\bigg)}} \\ \end{gathered} \\ [/tex]

4. Area of a triangle

Let A(x₁, y₁), B(x₂, y₂) and C(x₃, y₃) be the vertices of a triangle, then the area of triangle is given by

[tex]\begin{gathered}\boxed{\tt{ Area =\dfrac{1}{2}\bigg|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)\bigg|}} \\ \end{gathered} \\ [/tex]

Step-by-step explanation:

Hope this helps you!!!!!!!