Question

Show that the line segment joining the points A(-5,8) and B(10,-4) is trisected by the co-ordinate axes.

Answer 1

Answer.....

AP=PQ=QB

AP:PB=1:2

here, P is the point.

STEPS...

$(1 \times 10 + 2 \times ( - 5) \sqrt{1 + 2 }$

$(1 \times ( - 4) + 2 \times 8 \sqrt[]{1 + 2}$

answer=(0,4)

P lies on the y-axis

Q lies on the x-axis

AQ:QB=2:1

$(2 \times 10 + 1 \times ( - 5) \sqrt[?]{ ]{2 + 1} }$

$\sqrt[?]{?} 2 \times ( - 4) + 1 \times 8 \sqrt{2 + 1}$

answer =5,0

A and B are the trisected by co-cordinate axis.

AQ=QB

Answer 2

Step-by-step explanation:

Given points are A(-5,8) and B(10,-4).

Let p(x,0) and Q(0,y) be the points on coordinate axes that trisects AB.

(i)

Let the ratio in which p divides AB be m:n

Now,

P(x,0), (x₁,y₁) = (-5,8) and (x₂,y₂) = (10,-4) and in the ratio of m:n.

⇒ 0 = (my₂ + ny₁)/(m + n)

⇒ 0 = m(-4) + n(8)/m + n

⇒ -4m + 8n = 0

⇒ -4m = -8n

⇒ 4m = 8n

⇒ (m/n) = (8/4)

⇒ m : n = 2 : 1

Thus, P divides AB in the ratio 2 : 1.

For P(x,0), m = 2, n = 1, (x₁,y₁) = (-5,8) and (x₂,y₂) = (10,-4).

⇒ x = (mx₂ + nx₁)/(m + n)

⇒ x = [2(10) + 5]/(3)

⇒ x = 5

Therefore, the coordinates of P are (5,0).

(ii)

Let the ratio in which Q divides AB be a : b.

Now, Q(0,y), (x₁,y₁) = (-5,8) and (x₂,y₂) = (10,-4) divides in the ratio a : b.

⇒ 0 = (mx₂ + nx₁)/(m + n)

⇒ 0 = a(10) + b(-5)/(a + b)

⇒ 0 = 10a - 5b

⇒ 10a = 5b

⇒ (a/b) = 5/10

⇒ a : b = 1 : 2

Thus, Q divides AB in the ratio 1 : 2.

For Q(0,y), m = 1, n = 2.

⇒ y = (my₂ + ny₁)/(m + n)

      = (-4 - 8)/3

       = 4

Therefore, the coordinates of Q are (4,0).

Now,

From (i)

AP/PB = 2/1

⇒ AP = 2PB

From (ii)

AQ/QB = 1/2

⇒ 2AQ = QB

Now,

(iii)

AP + PB = AB

2PB + PB = AB

3PB = AB

PB = (1/3) AB

(iv)

AQ + QB = AB

⇒ AQ + 2AQ = AB

⇒ 3AQ = AB

⇒ AQ = (1/3)AB

From (iii),(iv)

PB = AQ.      ----- (v)

Also,

QB = QP + PB

⇒ 2AQ = QP + AQ

⇒ AQ = QP           ---------- (vi)

From (v),(vi)

AQ = QP = PB

Hope it helps!