Question

find the ratio in which of the line 4x+y=13 divides the line segment by the two points (1,6)and(6,1),​

Answer 1

Answer:

1 : 4 internally

Step-by-step explanation:

Let line CD: 4x + y = 13 divides the line segment by the two points A(1,6) and B(6,1) in the ratio m:n at point P(x,y)

m:n = $\frac{m}{n}$:1 = k = k:1 (Let)

So,

equation of line AB is given by

y - 6 = $\frac{1 - 6}{6 - 1}$ (x - 1)

y - 6 = -(x - 1)

y - 6 = - x + 1

x + y - 7 = 0

P is the intersection of lines AB and CD

So, by solving 4x + y - 13 = 0 and x + y -7 = 0, we get

P(x,y) = P(2,5)

Now,

by internal section formula

    (2,5) = ( $\frac{k.6 + k.1}{k + 1}$ , $\frac{k.1 + 1.6}{k + 1}$ )

or, (2,5) = ( $\frac{6k + 1}{k + 1}$ , $\frac{k + 6}{k + 1}$ )

Comparing corresponding elements,

    2 = ( $\frac{6k + 1}{k + 1}$ )

or, 2(k + 1) = 6k + 1

or, 2k + 2 = 6k + 1

or, 1 = 4k

or, k = $\frac{1}{4}$

or, $\frac{m}{n}$ = $\frac{1}{4}$

∴ m:n = 1:4 internally