Question

. The line segment joining the points A(-2, -3) and B( 2, -1) is divided by the y axis in the ratio
i. 1 : 2 ii. 2 : 1 iii. 1 : 1 iv. 1 : 3​​

Answer 1

$\large\underline{\sf{Solution-}}$

Given,

A line segment joining the points A(-2, -3) and B( 2, -1) is divided by the y axis.

Let assume that

y - axis divides the line segment joining the points A(-2, -3) and B( 2, -1) in the ratio k : 1 and let coordinates of point of intersection of line AB with y - axis be C (0, y).

So, it means C (0, y) divides the line segment joining the points A(-2, -3) and B( 2, -1) in the ratio k : 1.

We know,

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 will be:

$\begin{gathered} \boxed{\tt{ (x, y) = \bigg(\dfrac{m_{1}x_{2}+m_{2}x_{1}}{m_{1}+m_{2}}, \dfrac{m_{1}y_{2}+m_{2}y_{1}}{m_{1}+m_{2}}\bigg)}} \\ \end{gathered}$

So, we have

$\rm \: x_1 = - 2$

$\rm \: x_2 = 2$

$\rm \: y_1 = - 3$

$\rm \: y_2 = - 1$

$\rm \: m_1 = k$

$\rm \: m_2 = 1$

So, on substituting the values, we get

$\rm \: (0, \: y) = \bigg(\dfrac{2k - 2}{k + 1} , \: \dfrac{ - k - 3}{k + 1} \bigg)$

So, on comparing x - coordinate, we get

$\rm \: \dfrac{2k - 2}{k + 1} = 0$

$\rm \: 2k - 2 = 0$

$\rm \: 2k = 2$

$\rm \: k = 1$

So, it implies

y - axis divides the line segment joining the points A(-2, -3) and B( 2, -1) in the ratio 1 : 1.

So, option (iii) is correct.

Remark :- Short Cut Trick

If a line segment joining the points A(x₁, y₁) and B(x₂, y₂) is divided by y - axis, then required ratio is - x₁ : x₂.

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Additional Information :-

1. 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

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

2. 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

$\begin{gathered} \boxed{\tt{ (x, y) = \bigg(\dfrac{m_{1}x_{2}+m_{2}x_{1}}{m_{1}+m_{2}}, \dfrac{m_{1}y_{2}+m_{2}y_{1}}{m_{1}+m_{2}}\bigg)}} \\ \end{gathered}$

3. 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

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

4. 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

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

5. 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

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

6. Condition for 3 points to be Collinear

Let A(x₁, y₁), B(x₂, y₂) and C(x₃, y₃) be the coordinates in cartesian plane, then points A, B and C are collinear, then

$\begin{gathered}\boxed{\tt{x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2) = 0}} \\ \end{gathered}$

Answer 2

Answer:

Bryophyllum is a group of plant species of the family Crassulaceae native to Madagascar. It is a section or subgenus within the genus Kalanchoe, and was formerly placed at the level of genus.

Step-by-step explanation:

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