Question

42. Find the ratio in which the line x - 3y = 0 divides the line segment joining the
points (-2,-5) and (6, 3). Find the co-ordinates of the point of intersection.
HOTS​

Answer 1

Solution :-

Let us assume that, the given line x - 3y = 0 intersect the line segment joining A(-2, -5) and B(6, 3) in the ratio k : 1 at point P(x,y).

Now the section formula says that, the point which divides the line joining two points (a ,b) and (c , d) in the ratio m : n is :- P(x , y) = (cm + an)/(m + n) , (dm + bn)/(m + n) .

So,

Putting values now, we get :-

→ Coordinates of P(x, y) = (6k - 2)/(k+1), (3k - 5)/(k+1)

Now, we have also given that, P lies on x - 3y = 0 .

→ x = 3y

Or,

→ (6k - 2)/(k + 1) = 3{(3k - 5)/(k + 1)}

(k + 1) will be cancel from both denominators,,

→ 6k - 2 = 9k - 15

→ 9k - 6k = 15 - 2

→ 3k = 13

→ k = (13/3)

Therefore,

→ Required Ratio = k : 1 = (13/3) : 1 = 13 : 3

Now,

→ Coordinates of P(x, y) = (6k - 2)/(k+1), (3k - 5)/(k+1)

Putting value of k = (13/3) , we get,

→ x = { 6*(13/3) - 2 } / (13/3 + 1)

→ x = ( 26 - 2 ) / {(13 + 3) / 3}

→ x = 24 * 3 / 16

→ x = (3 * 3)/2

→ x = (9/2)

and,

→ y = { 3*(13/3) - 5 } / (13/3 + 1)

→ y = ( 13 - 5 ) / {(13 + 3) / 3}

→ y = 8 * 3 / 16

→ y = (3/2)

Hence, Coordinates of P are (9/2, 3/2) .

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