Question

In a GPS, The lines that run east-west are known as lines of latitude, and the lines running north-south are known as lines of longitude. The latitude and the longitude of a place are its coordinates and the distance formula is used to find the distance between two places. The distance between two parallel lines is approximately 150 km. A family from Uttar Pradesh planned a round trip from Lucknow (L) to Puri (P) via Bhuj (B) and Nashik (N) as shown in the given figure below.

Based on the above information answer the following questions using the coordinate geometry.

I. Find the distance between Lucknow (L) to Bhuj(B).

II. If Kota (K), internally divide the line segment joining Lucknow (L) to Bhuj (B) into 3 : 2 then find the coordinate of Kota (K).

III. Name the type of triangle formed by the places Lucknow (L), Nashik (N) and Puri (P)

or

Find a place (point) on the longitude (y-axis) which is equidistant from the points Lucknow (L) and Puri (P).

Answer 1

I. LB = \(\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\)

⇒ LB = \(\sqrt{(0 -5)^2+(7 - 10)^2}\)

LB = \(\sqrt{(5)^2 + (3)^2}\)

⇒ LB = \(\sqrt{25 + 9}\)

LB = \(\sqrt{34}\)

Hence the distance is 150√34 km.

II. Coordinate of Kota (K) is 

\(\left(\frac{3\times 5 + 2 \times 0}{3 + 2}, \frac{3\times 7 + 2\times 10}{3 + 2}\right)\)

\(= \left(\frac{15 + 0}{5}, \frac{21 + 20}{5}\right)\)

\(= (3, \frac{41}5)\)

III. L(5, 10), N(2, 6), P(8, 6)

LN = \(\sqrt{(2 - 5)^2 + (6 - 10)^2}\)

\(= \sqrt{(3)^2 + (4)^2}\)

\(= \sqrt{9 + 16}\)

\(= \sqrt{25}\)

\(= 5\)

NP = \(\sqrt{(8 - 2)^2 + (6 - 6)^2}\)

\(= \sqrt{(4)^2 + (0)^2}\)

\(= 4\)

PL = \(\sqrt{(8 - 5)^2 + (6 - 10)^2}\)

\(= \sqrt{(3)^2 + (4)^2}\)

⇒ LB = \(\sqrt{9 + 16}\)

\(= \sqrt{25}\)

= 5

As LN = PL ≠ NP, So ∆LNP is an isosceles triangle.

OR

Let A(0, b) be a point on the y-axis then AL = AP

⇒ \(\sqrt{(5 - 0)^2 + (10 - b)^2} = \sqrt{(8 - 0)^2 + (6 - b)^2}\)

⇒ (5)² + (10 − b)² = (8)² + (6 − b)²

⇒ 25 + 100 − 20b + b² = 64 + 36 − 12b + b²

⇒ 8b = 25

⇒ b = \(\frac{25}8\)

So, the coordinate on y axis is \((0, \frac{25}8)\).

Answer 2

(i)

 

(ii)

Coordinate of Kota (K) is

= \((\frac{3\times 0+2\times 5}{3+2} ,\frac{3\times 7+2\times 10}{3+2})\)

= \((\frac{10}{5},\frac{21+20}{5})=(2,\frac{41}{5})\) 

(iii) 

L(5, 10), N(2,6), P(8,6)

\(LN = \sqrt{(2-5)^2 +(6-10)^2}\)

\(=\sqrt{(3)^2 +(4)^2}=\sqrt{9+16}=\sqrt{25}=5\)

NP = \(\sqrt{(8-2)^2 +(6-6)^2}=\sqrt{(6)^2 +(0)^2}=4\)

PL = \(\sqrt{(8-5)^2 +(6-10)^2}=\sqrt{(3)^2 +(4)^2}\) 

⇒ PL = \(\sqrt{9+16}=\sqrt{25}=5\)

as LN = PL ≠ NP, so ΔLNP is an isosceles triangle.

or

Let A (0, b ) be a point on the y – axis then AL = AP

So, the coordinate on y axis is (0,25/8)