Question

The endpoints of AB are A(1,4) and B(6,-1).

Answer 1

Answer:

C is $(3,2)$

D is $(2.2,2.8)$

Step-by-step explanation:

(i) We need to find the coordinate of point C, which divides A(1,4) and B(6,-1) in the ratio 2:3

We know that the coordinate of a point $(x,y)$ dividing a line segment joining $(x_{1},y_{1}) \:\text{and} (x_{2},y_{2})$ in the ratio m:n is given by

$x=\frac{mx_{2}+nx_{1}}{m+n}$, $y=\frac{my_{2}+ny_{1}}{m+n}$

Let the coordinate of point C be $(x,y)$

Here, the ratio is 2:3

So, $x=\frac{2(6)+3(1)}{2+3}$

$x=\frac{15}{5}$

$x=3$

$y=\frac{2(-1)+3(4)}{2+3}$

$y=\frac{-2+12}{5}$

$y=\frac{10}{5}$

$y=2$

Hence, coordinate of C is $(3,2)$

(ii) We need to find the coordinate of point D, which divides A(1,4) and C(3,2) in the ratio 3:2

We know that the coordinate of a point $(x,y)$ dividing a line segment joining $(x_{1},y_{1}) \:\text{and} (x_{2},y_{2})$ in the ratio m:n is given by

$x=\frac{mx_{2}+nx_{1}}{m+n}$, $y=\frac{my_{2}+ny_{1}}{m+n}$

Let the coordinate of point D be $(x_{3},y_{3})$

Here, the ratio is 2:3

So, $x_{3}=\frac{3(3)+2(1)}{3+2}$

$x_{3}=\frac{11}{5}$

$x_{3}=2.2$

$y_{3=\frac{3(2)+2(4)}{3+2}$

$y_{3=\frac{6+8}{5}$

$y_{3=\frac{14}{5}$

$y_{3=2.8$

Hence, coordinate of D is $(2.2,2.8)$

Answer 2

Answer:

C(3, 2)

D(2.2, 2.8)

Step-by-step explanation:

If point $M(x,y)$ divides the segment AB with endpoints $A(x_1,y_1)$ and $B(x_2,y_2)$ in the ratio $m:n,$ then its coordinates are

$x=\dfrac{n\cdot x_1+m\cdot x_2}{m+n}\\ \\y=\dfrac{n\cdot y_1+m\cdot y_2}{m+n}$

You are given A(1,4) and B(6,-1).

1. If point C divides AB in the ratio 2 : 3, the coordinates of point C are

$x=\dfrac{3\cdot 1+2\cdot 6}{2+3}=\dfrac{3+12}{5}=\dfrac{15}{5}=3\\ \\y=\dfrac{3\cdot 4+2\cdot (-1)}{2+3}=\dfrac{12-2}{5}=\dfrac{10}{5}=2$

C(3, 2)

2. If point D divides AC in the ratio 3 : 2, the coordinates of point D are

$x=\dfrac{2\cdot 1+3\cdot 3}{3+2}=\dfrac{2+9}{5}=\dfrac{11}{5}=2.2\\ \\y=\dfrac{2\cdot 4+3\cdot 2}{3+2}=\dfrac{8+6}{5}=\dfrac{14}{5}=2.8$

D(2.2, 2.8)