Question

Find the ratio in which (4,5) divides the join (2,3) and (7,8)

Answer 1

Let the point P (4,5) divides the segment A(2,3) and (7,8) in the ratio of k:1.

The Division formula , i.e if a point P(x,y) divides (a,b) and (c,d) in m:n, then

x = $\frac{mc+na}{m+n}$ and y = $\frac{md+nb}{m+n}$

Applying the formula,

$4= \frac{7 k +2}{k+1}$

→ 4(k +1)= 7 k+ 2

→ 4 k +4 =7 k +2

→ 7 k - 4 k= 4-2

→ 3 k = 2

→ k =2/3

So, (4,5) divides the join of (2,3) and (7,8) in the ratio of 2:3.

Answer 2

Answer:

(4,5) divides the join (2,3) and (7,8) in the ratio 2:3.

Concept:

Section formula.

Suppose a point P (x,y) divides the join of A(x₁, y₁) and B (x₂, y₂) in the ratio m:n, then the point P is given by section formula as follows -

P(x,y) = $(\frac{mx_2 + nx_1}{m+n} ,\frac{my_2 + ny_1}{m + n} )$

Find:

The ratio in which (4,5) divides the join (2,3) and (7,8).

Solution:

Let point P(4,5) divide the join of A(2,3) and B(7,8) in the ratio m:n.

Let m:n = k

∴ The ratio becomes k:1.

Now, the coordinates of point P are given by,

P(4,5) = $P(\frac{k(7) + 1(2)}{k + 1} ,\frac{k(8) + 1(3)}{k + 1} )$

$(4, 5) = (\frac{7k + 2}{k + 1} , \frac{8k + 3}{ k + 1} )$

Comparing both sides we get,

$4 = \frac{7k + 2}{k + 1} ---------- (i)\\5 = \frac{8k + 3}{k + 1} ----------(ii)$

On solving (i), we have

4(k + 1) = 7k + 2

4k + 4 = 7k + 2

4 - 2 = 7k - 4k

2 = 3k

3k = 2

k = 2/3

m:n = 2:3

Hence, the required ratio is 2 : 3.

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