Question

In what ratio does the point (3, a) divide the join of (1, 7) & (6, −3)? Also, find a.

Answer 1

The ratio is 2:3. The value of a is 3.

Given:

Point (3, a) divide the join of (1, 7) and (6, −3).

To Find:

In what ratio does point (3, a) divide the join of (1, 7) & (6, −3)?

The value of a =?

Solution:

Let the ratio in which the point (3, a) divides the join of (1, 7) & (6, −3) be k:1.

The section formula will be used here.

If a point P(x,y) divides the line AB with the coordinates of A and B as (x₁,y₁) and (x₂,y₂) respectively in the ratio m:n, then

$x = \frac{mx_2 +nx_1}{m+n}$ and $y = \frac{my_2 +ny_1}{m+n}$

Using the same formula, we can find the ratio as well as the value of a.

$3 = \frac{6k +1}{k+1}$ -------------(1) and

$a = \frac{-3k +7}{k+1}$ ------------(2)

Solving the first question $3 = \frac{6k +1}{k+1}$, we get

3k + 3 = 6k + 1

3 - 1 = 6k - 3k

2 = 3k

k = 2/3

k : 1 = $\frac{2}{3}: 1$ = 2 : 3

Putting the value of k in equation (2), we get

$a = \frac{-3k +7}{k+1}$

$a = \frac{-3(\frac{2}{3}) +7}{\frac{2}{3} +1}$

$a = \frac{-2 +7}{\frac{5}{3} }$

$a = \frac{5}{\frac{5}{3} }$

a = 3

The ratio is 2:3. The value of a is 3.

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