Question

5 - The origin O, B (-6, 9) and C (12, -3) are vertices of triangle OBC. Point P divides OB in the ratio 1: 2 and point Q divides OC in the ratio 1:2. Find the 1 co-ordinates of points P and Q. Also, show that : PQ = BC. 3 AB​

Answer 1

Step-by-step explanation:

Corrected Question:-

Show that PQ = BC/3

Given :-

The origin O, B (-6, 9) and C (12, -3) are vertices of triangle OBC.

Point P divides OB in the ratio 1: 2

Point Q divides OC in the ratio 1:2.

To find :-

1) The co-ordinates of points P and Q.

2) PQ = BC/ 3

Solution :-

Given points are Origin , B(-6,9) and C(12,-3)

The coordinates of the origin = O(0,0)

Given that

O, B and C are the vertices of the ∆ OBC

P divides OB in the ratio 1:2

Let (x1, y1) = O(0,0) => x1 = 0 and y1 = 0

Let (x2, y2) = B(-6,9) => x2 = -6 and y2 = 9

Let m1 :m2 = 1:2 => m1 = 1 and m2 = 2

We know that

The coordinates of the point P(x,y) which divides the linesegment joining the points (x1, y1) and (x2, y2) in the ratio m1:m2 internally is

({m1x2+m2x1}/(m1+m2) , {m1y2+m2y1}/(m1+m2))

Now,

P(x,y) = ({(1×-6)+(2×0})/(1+2) , {(1×9)+(2×0)}/(1+2))

=> P(x,y) = ({-6+0}/3 , {9+0}/3)

=> P(x,y) = (-6/3 , 9/3 )

=> P(x,y) = (-2,3)

Therefore, The coordinates of P = (-2,3)

And

Q divides OC in the ratio 1:2

Let (x1, y1) = O(0,0) => x1 = 0 and y1 = 0

Let (x2, y2) = C(12,-3) => x2 = 12 and y2 = -3

Let m1 :m2 = 1:2 => m1 = 1 and m2 = 2

We know that

The coordinates of the point Q(x,y) which divides the linesegment joining the points (x1, y1) and (x2, y2) in the ratio m1:m2 internally is

({m1x2+m2x1}/(m1+m2) , {m1y2+m2y1}/(m1+m2))

Now,

Q(x,y) = ({(1×12)+(2×0})/(1+2) , {(1×-3)+(2×0)}/(1+2))

=> Q(x,y) = ({12+0}/3 , {-3+0}/3)

=> Q(x,y) = (12/3 , -3/3 )

=> Q(x,y) = (4,-1)

Therefore, The coordinates of Q = (4,-1)

Finding PQ :-

Let (x1, y1) = (-2,3) => x1 = -2 and y1 = 3

Let (x2, y2) = (4,-1) => x2 = 4 and y2 = -1

We know that

The distance between two points (x1, y1) and (x2, y2) is √[(x2-x1)²+(y2-y1)²] units

The distance between two points P and Q

=> PQ = √[(4-(-2))²+(-1-3)²]

=> PQ = √(4+2)²+(-4)²]

=> PQ = √[6²+(-4)²]

=> PQ = √(36+16)

=> PQ = √52

=> PQ = √(4×13)

=> PQ = 2√13 units -----------(1)

Finding BC :-

Let (x1, y1) = (-6,9) => x1 = -6 and y1 = 9

Let (x2, y2) = (12,-3) => x2 = 12 and y2 = -3

We know that

The distance between two points (x1, y1) and (x2, y2) is √[(x2-x1)²+(y2-y1)²] units

The distance between two points B and C

=> BC = √[(12-(-6))²+(-3-9)²]

=> BC = √(12+6)²+(-12)²]

=> BC = √[18²+(-12)²]

=> BC = √(324+144)

=> BC = √468

=> BC = √(36×13)

=> BC = 6√13 units

Now, BC /3 = 6√13/3 = 2√13 units----------(2)

Therefore, From (1)&(2)

PQ = BC/3

Answer :-

1) The coordinates of the point P = (-2,3)

2) The coordinates of the point Q = (4,-1)

3) PQ = BC/3

Used formulae

The distance between two points (x1, y1)

and (x2, y2) is √[(x2-x1)²+(y2-y1)²] units

→ The coordinates of the point P(x,y) which divides the linesegment joining the points (x1, y1) and (x2, y2) in the ratio m1:m2 internally is

({m1x2+m2x1}/(m1+m2) , {m1y2+m2y1}/(m1+m2))

Answer 2

Step-by-step explanation:

$\text{ \color{maroon} For point P\( \color{black} \tt: m_{1}: m_{2}=1: 2,\left(x_{1}, y_{1}\right)=(0,0) \)}$

$\[ \begin{aligned} \tt \text { and }\left(x_{2}, y_{2}\right)=(-6,9) & \\ \\ \tt\therefore P & \tt=\left(\frac{m_{1} x_{2}+m_{2} x_{1}}{m_{1}+m_{2}}, \frac{m_{1} y_{2}+m_{2} y_{1}}{m_{1}+m_{2}}\right) \\ \\ & \tt=\left(\frac{1 \times-6+2 \times 0}{1+2}, \frac{1 \times 9+2 \times 0}{1+2}\right) \\ \\ & \color{maroon}\tt=(-2,3) \end{aligned} \]$

$\text{ \color{brown}For point \( \tt Q \color{black}: m_{1}: m_{2}=1: 2,\left(x_{1}, y_{1}\right)=(0,0) \)\color{black} and \(\color{black} \tt \left(x_{2}, y_{2}\right)=(12,-3) \)}$

$\tt \therefore \quad Q =\left(\dfrac{1 \times 12+2 \times 0}{1+2}, \dfrac{1 \times-3+2 \times 0}{1+2}\right) \color{brown}=(4,-1)$

$\text{ Now \( \bf PQ = \) Distance between \( \bf P (-2,3) \) and \( \bf Q (4,-1) \)}$

$\bf \[ =\sqrt{(4+2)^{2}+(-1-3)^{2}}=\sqrt{36+16}=\sqrt{52}=2 \sqrt{13} \]$

and,

$\bf \quad BC =\sqrt{(12+6)^{2}+(-3-9)^{2}}=\sqrt{324+144}=\sqrt{468}=6 \sqrt{13}$

$\text{\( \bf PQ =2 \sqrt{13} \) and \( \bf BC =6 \sqrt{13} \Rightarrow \quad \color{brown} PQ =\dfrac{1}{3} BC \)}$