Question

URGENT ! Pls answer 2nd question ​

Answer 1

Topic :-

Straight Line

Given :-

P(-1, 4) and Q(11, -8) divide AB harmonically in the ratio 2 : 3.

To Find :-

Coordinates of A and B.

Concept Used :-

Two points dividing a line segment harmonically means one point is dividing it internally and other is dividing it externally.

Section Formula

The coordinates of point dividing a line segment in ratio m : n is given by :-

For Internal Division

$\sf{(x,y) \equiv \left (\dfrac{mx_2+nx_1}{m+n},\dfrac{my_2+ny_1}{m+n} \right )}$

For External Division

$\sf{(x,y) \equiv \left (\dfrac{mx_2-nx_1}{m-n},\dfrac{my_2-ny_1}{m-n} \right )}$

Solution :-

Assume coordinates of A and B as :-

$\sf {A (x_1,y_1)}$

$\sf {B (x_2,y_2)}$

Let us assume that point P divides the line segment AB internally.

P(-1, 4)

m : n is 2 : 3

Applying formula for internal division,

$\sf{(-1,4) \equiv \left (\dfrac{2x_2+3x_1}{2+3},\dfrac{2y_2+3y_1}{2+3} \right )}$

$\sf{(-1,4) \equiv \left (\dfrac{2x_2+3x_1}{5},\dfrac{2y_2+3y_1}{5} \right )}$

$\sf {-1=\dfrac{2x_2+3x_1}{5}}$

$\sf{-5=2x_2+3x_1}$

$\sf {4=\dfrac{2y_2+3y_1}{5}}$

$\sf{20=2y_2+3y_1}$

Applying formula for external division,

$\sf{(11,-8) \equiv \left (\dfrac{2x_2-3x_1}{2-3},\dfrac{2y_2-3y_1}{2-3} \right )}$

$\sf{(11,-8) \equiv \left (\dfrac{2x_2-3x_1}{-1},\dfrac{2y_2-3y_1}{-1} \right )}$

$\sf{(11,-8) \equiv \left ({3x_1-2x_2},{3y_1-2y_2} \right )}$

$\sf{11 = 3x_1-2x_2}$

$\sf{-8= 3y_1-2y_2}$

Now, solve the equations,

$\sf{2x_2+3x_1+3x_1-2x_2=-5+11}$

$\sf{6x_1=6}$

$\sf{x_1=1}$

$\sf{11 = 3x_1-2x_2}$

$\sf{11 = 3(1)-2x_2}$

$\sf {2x_2=3-11}$

$\sf {2x_2=-8}$

$\sf {x_2=-4}$

$\sf{2y_2+3y_1+3y_1-2y_2=20-8}$

$\sf{6y_1=12}$

$\sf{y_1=2}$

$\sf{20=2y_2+3y_1}$

$\sf{20=2y_2+3(2)}$

$\sf{2y_2=20-6}$

$\sf{2y_2=14}$

$\sf{y_2=7}$

So,

$\sf {A (x_1,y_1)\equiv A (1,2)}$

$\sf {B (x_2,y_2)\equiv B (-4,7)}$

Answer :-

The coordinates of A and B are in order (1, 2), (-4, 7) which is option B.

Answer 2

Answer:

Topic :-

Straight Line

Given :-

P(-1, 4) and Q(11, -8) divide AB harmonically in the ratio 2 : 3.

To Find :-

Coordinates of A and B.

Concept Used :-

Two points dividing a line segment harmonically means one point is dividing it internally and other is dividing it externally.

Section Formula

The coordinates of point dividing a line segment in ratio m : n is given by :-

For Internal Division

\sf{(x,y) \equiv \left (\dfrac{mx_2+nx_1}{m+n},\dfrac{my_2+ny_1}{m+n} \right )}(x,y)≡(m+nmx2+nx1,m+nmy2+ny1)

For External Division

\sf{(x,y) \equiv \left (\dfrac{mx_2-nx_1}{m-n},\dfrac{my_2-ny_1}{m-n} \right )}(x,y)≡(m−nmx2−nx1,m−nmy2−ny1)

Solution :-

Assume coordinates of A and B as :-

\sf {A (x_1,y_1)}A(x1,y1)

\sf {B (x_2,y_2)}B(x2,y2)

Let us assume that point P divides the line segment AB internally.

P(-1, 4)

m : n is 2 : 3

Applying formula for internal division,

\sf{(-1,4) \equiv \left (\dfrac{2x_2+3x_1}{2+3},\dfrac{2y_2+3y_1}{2+3} \right )}(−1,4)≡(2+32x2+3x1,2+32y2+3y1)

\sf{(-1,4) \equiv \left (\dfrac{2x_2+3x_1}{5},\dfrac{2y_2+3y_1}{5} \right )}(−1,4)≡(52x2+3x1,52y2+3y1)

\sf {-1=\dfrac{2x_2+3x_1}{5}}−1=52x2+3x1

\sf{-5=2x_2+3x_1}−5=2x2+3x1

\sf {4=\dfrac{2y_2+3y_1}{5}}4=52y2+3y1

\sf{20=2y_2+3y_1}20=2y2+3y1

Applying formula for external division,

\sf{(11,-8) \equiv \left (\dfrac{2x_2-3x_1}{2-3},\dfrac{2y_2-3y_1}{2-3} \right )}(11,−8)≡(2−32x2−3x1,2−32y2−3y1)

\sf{(11,-8) \equiv \left (\dfrac{2x_2-3x_1}{-1},\dfrac{2y_2-3y_1}{-1} \right )}(11,−8)≡(−12x2−3x1,−12y2−3y1)

\sf{(11,-8) \equiv \left ({3x_1-2x_2},{3y_1-2y_2} \right )}(11,−8)≡(3x1−2x2,3y1−2y2)

\sf{11 = 3x_1-2x_2}11=3x1−2x2

\sf{-8= 3y_1-2y_2}−8=3y1−2y