Question

Internal
External
No.
Points
Distance
Mid
Point
Point
Ratio
Point
Ratio
(i)
(3, 4), (5,5)
2:3
2:3​

Answer 1

Step-by-step explanation:

Internal

External

Are your ans

Answer 2

Answer:

Let (x₁, y₁) and (x₂, y₂) be the cartesian co-ordinates of the points P and Q respectively referred to rectangular co-ordinate axes OX and OY and the point R divides the line-segment PQ internally in a given ratio m : n (say), i.e., PR : RQ = m : n. We are to find the co-ordinates of R.

Let, (x, y) be the required co-ordinate of R . From P, Q and R, draw PL, QM and RN perpendiculars on OX. Again, draw PT parallel to OX to cut RN at S and QM at T.

Then,

PS = LN = ON - OL = x – x₁;

PT = LM = OM – OL = x₂ - x₁;

RS = RN – SN = RN – PL = y - y₁;

and QT = QM – TM = QM – PL = y₂ – y₁

Again, PR/RQ = m/n

or, RQ/PR = n/m

or, RQ/PR + 1 = n/m + 1

or, (RQ + PR/PR) = (m + n)/m

o, PQ/PR = (m + n)/m

Now, by construction, the triangles PRS and PQT are similar; hence,

PS/PT = RS/QT = PR/PQ

Taking, PS/PT = PR/PQ we get,

(x - x₁)/(x₂ - x₁) = m/(m + n)

or, x (m + n) – x₁ (m + n) = mx₂ – mx₁

or, x ( m + n) = mx₂ - mx₁ + m x₁ + nx₁ = mx₂ + nx₁

Therefore, x = (mx2 + nx1)/(m + n)

Again, taking RS/QT = PR/PQ we get,

(y - y₁)/(y₂ - y₁) = m/(m + n)

or, ( m + n) y - ( m + n) y₁ = my₂ – my₁

or, ( m+ n)y = my₂ – my₁ + my₁ + ny₁ = my₂ + ny₁

Therefore, y = (my₂ + ny₁)/(m + n)

Therefore, the required co-ordinates of the point R are

((mx₂ + nx₁)/(m + n), (my₂ + ny₁)/(m + n))

Step-by-step explanation:

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