Question

Find the coordinates of the point which Divides the Join of Points.(3,4)and(5,6)and the ratio 1:2.
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Answer 1

Answer:

The coordinates of the point dividing the line joining the given points are

$\displaystyle{\boxed{\red{\sf\:(\:x\:,\:y\:)\:=\:\left(\:\dfrac{11}{3}\:,\:\dfrac{14}{3}\:\right)\:}}}$

Step-by-step-explanation:

We have given the coordinates of two points.

Let the two points be A, B.

• A ≡ ( 3, 4 ) ≡ ( x₁, y₁ )
• B ≡ ( 5, 6 ) ≡ ( x₂, y₂ )

We have to find the coordinates of the point which divides the line joining A & B in the ratio 1 : 2.

Let that point be C.

• C ≡ ( x, y )
• Ratio = m : n = 1 : 2

Now, by section formula,

$\displaystyle{\pink{\sf\:x\:=\:\dfrac{mx_2\:+\:nx_1}{m\:+\:n}}}$

$\displaystyle{\implies\sf\:x\:=\:\dfrac{1\:\times\:5\:+\:2\:\times\:3}{1\:+\:2}}$

$\displaystyle{\implies\sf\:x\:=\:\dfrac{5\:+\:6}{3}}$

$\displaystyle{\implies\boxed{\blue{\sf\:x\:=\:\dfrac{11}{3}}}}$

Now, by section formula,

$\displaystyle{\pink{\sf\:y\:=\:\dfrac{my_2\:+\:ny{_1}}{m\:+\:n}}}$

$\displaystyle{\implies\sf\:y\:=\:\dfrac{1\:\times\:6\:+\:2\:\times\:4}{1\:+\:2}}$

$\displaystyle{\implies\sf\:y\:=\:\dfrac{6\:+\:8}{3}}$

$\displaystyle{\implies\boxed{\green{\sf\:y\:=\:\dfrac{14}{3}}}}$

∴ The coordinates of the point dividing the line joining the given points are

$\displaystyle{\underline{\boxed{\red{\sf\:(\:x\:,\:y\:)\:=\:\left(\:\dfrac{11}{3}\:,\:\dfrac{14}{3}\:\right)\:}}}}$

Answer 2

Given :

• Line segment, coordinates of its ends = (3,4) and (5,6)

To Find :

• Coordinates of point which divides the line segment in ratio 1:2

Solution :

• Let ends of line segment be P(3,4) and Q(5,6)
• And point dividing it in ratio 1:2 be R(x,y)

We know that according to section formula,

${\underline{\boxed{\bf{(x,y) = \Bigg(\dfrac{m_{1}x_{2} + m_{2}x_{1}}{m_{1} + m_{2}},\dfrac{m_{1}y_{2} + m_{2}y_{1}}{m_{1} + m_{2}}\Bigg)\:\:\:}}}}$

Finding first coordinate x,

$\\ :\implies\:\sf x = \dfrac{m_{1}x_{2} + m_{2}x_{1}}{m_{1} + m_{2}}$

We have,

• x₁ = 3
• x₂ = 5
• m₁ = 1
• m₂ = 2

Putting all values,

$\\ :\implies\:\sf x = \dfrac{1(5) + 2(3)}{1 + 2}$

$\\ :\implies\:\sf x = \dfrac{(1\:\times\:5) + (2\:\times\:3)}{3}$

$\\ :\implies\:\sf x = \dfrac{5 + 6}{3}$

$\\ :\implies\:\sf\red{x = \dfrac{11}{3}}$

Finding second coordinate y,

$\\ :\implies\:\sf y = \dfrac{m_{1}y_{2} + m_{2}y_{1}}{m_{1} + m_{2}}$

We have,

• y₁ = 4
• y₂ = 6
• m₁ = 1
• m₂ = 2

Putting all values,

$\\ :\implies\:\sf y = \dfrac{1(6) + 2(4)}{1 + 2}$

$\\ :\implies\:\sf y = \dfrac{(1\:\times\:6) + (2\:\times\:4)}{3}$

$\\ :\implies\:\sf y = \dfrac{6 + 8}{3}$

$\\ :\implies\:\sf\red{y = \dfrac{14}{3}}$

Hence, required coordinates (x,y) = (11/3, 14/3).

M O R EㅤT OㅤK N O W

$\:$

• The distance between $\bf A(x_{1},\:y_{1})$ and $\bf B(x_{2},\:y_{2})$ is ::

$\bf \sqrt{\Big(x_{2} - x_{1}\Big)^2 + \Big(y_{2} - y_{1}\Big)^2}$

$\:$

• Distance of a point P(x, y) from the origin is ::

$\bf \sqrt{x^2 + y^2}$

$\:$

• The coordinates of the point P(x, y) which divides the line segment joining the points $\bf A(x_{1},\:y_{1})$ and $\bf B(x_{2},\:y_{2})$ internally in the ratio $\bf m_{1}\::\:m_{2}$ are ::

$\bf \Bigg(\dfrac{m_{1}x_{2} + m_{2}x_{1}}{m_{1} + m_{2}},\:\dfrac{m_{1}y_{2} + m_{2}y_{1}}{m_{1} + m_{2}} \Bigg)$

$\:$

• The mid - point of the line segment joining the points $\bf A(x_{1},\:y_{1})$ and $\bf B(x_{2},\:y_{2})$ is ::

$\bf \Bigg( \dfrac{x_{1} + x_{2}}{2},\:\dfrac{y_{1} + y_{2}}{2}\Bigg)$

$\:$

• The area of triangle formed by points $\bf (x_{1},\:y_{1})$, $\bf (x_{2},\:y_{2})$ and $\bf (x_{3},\:y_{3})$ is the numerical value of the expression ::

$\small\bf\dfrac{1}{2}\Big|x_{1}\big(y_{2} - y_{3}\big) + x_{2}\big(y_{3} - y_{1}\big) + x_{3}\big(y_{1} - y_{2}\big)\Big|$

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