**KSEEB SSLC Class 10 Maths Solutions Chapter 7 Coordinate Geometry Ex 7.2** are part of KSEEB SSLC Class 10 Maths Solutions. Here we have given Karnataka SSLC Class 10 Maths Solutions Chapter 7 Coordinate Geometry Exercise 7.2.

## Karnataka SSLC Class 10 Maths Solutions Chapter 7 Coordinate Geometry Exercise 7.2

Question 1.

Find the coordinates of the point which divides the join of (-1, 7) and (4, -3) in the ratio 2 : 3.

Solution:

Let P(x, y) is the required point.

P(x, y) divides the points A(-1, 7),

B (4, -3) in the ratio m_{1} : m_{2} = 2 : 3.

As per Section Formula, coordinates of P(x, y) are

Question 2.

Find the coordinates of the points of trisection of the line segment joining (4, -1) and (-2, -3).

Solution:

Let the given points be A(4, -1) and B(-2, -3)

Let the points P and Q trisects AB.

i.e., AP = PQ = QB

i.e., P divides AB in the ratio of 1 : 2 and Q divides AB in the ratio of 2 : 1

Let the coordinates of P be (x, y)

Question 3.

To conduct Sports Day activities, in your rectangular shaped school ground ABCD, lines have been drawn with chalk powder at a distance of 1 m each. 100 flower pots have been placed at a distance of 1m from each other along AD, as shown in the following figure.

Niharika runs \(\frac{1}{4}\) th the distance of AD on then 2nd line and posts a green flag

Preet runs \(\frac{1}{5}\) th the distance AD on the 5 eighthline and posts a red flag. What is the distance between both the flags ? If Rashmi has to post a blue flag exactly halfway between the line segment joining the two flags, where should she post her flag?

Solution:

Origin, A(0, 0)

AB along x – axis

AD along y – axis

In the parallel line AD in 2^{nd} row, Rashmi sees green flag at the distance of \(\frac{1}{4}\) .

∴ AD = \(\frac{1}{4}\) × 100 = 25 m.

∴ Coordinates of green flag are G(2, 25)

Similarly we can find out coordinates of Red flag as R(8, 20).

∴ Distance between G and R is

It means blue flag is in 5th line and parallel to AD at the distance of 22.5 m.

Question 4.

Find the ratio in which the line segment joining the points (-3, 10) and (6, -8) is divided by (-1, 6).

Solution:

Let the given points are A(-3, 10) and B(6, -8).

Let the point P(-1, 6) divides AB in the ratio m_{1} : m_{2}.

∴ Using the section formula, we have

⇒ -1(m_{1} + m_{2}) = 6m_{1} – 3m_{2}

and 6(m_{1} + m_{2}) = – 8m_{1} + 10m_{2}

⇒ -m_{1} – m_{2} – 6m_{1} + 3m_{2} = 0

and 6m_{1} + 6m_{2} + 8m_{1} – 10m_{2} = 0

⇒ -7m_{1} + 2m_{2} = 0 and 14m_{1} – 4m_{2} = 0 or 7m_{1} – 1m_{2} = 0

⇒ \(\frac{m_{1}}{m_{2}}=\frac{2}{7}\) and \(\frac{m_{1}}{m_{2}}=\frac{2}{7}\)

⇒ 2m_{2} = 7m_{1} and 7m_{1} = 2m_{2}

⇒ m_{1 }: m_{2} = 2 : 7 and m_{1} : m_{2} = 2 : 7

Thus, the required ratio is 2 : 7.

Question 5.

Find the ratio in whcih the line segment joining A(1, -5) and B(-4, 5) is divided by the x-axis. Also find the coordinates of the point of division.

Solution:

Let the line segment joining A and B points divides x-axis in the ratio k : 1.

If coordinates of y, P(x, 0) compared,

\(\frac{k \times(5)+1 \times(-5)}{k+1}=0\)

5k – 5 = 0

∴ k = 1

∴ Ratio will be k : 1 = 1 : 1

∴ Point P(x, 0) is the mid-point of line segment AB.

∴ x = \(\frac{1-4}{2}=\frac{-3}{2}\)

∴ Dividing point, P(\(-\frac{3}{2}\), 1)

Question 6.

If (1, 2), (4, y), (x, 6) and (3,5) are the vertices of a parallelogram taken in order, find x and y.

Solution:

Let the given points are A( 1, 2), B(4, y), C(x, 6) and D(3, 5)

Since, the diagonals of a parallelogram bisect each other.

∴ The coordinates of P are :

∴ The required values of x and y are 6 and 3 respectively.

Question 7.

Find the coordinates of point A, where AB is the diameter of a circle whose centre is (2, -3) and B is (1, 4).

Solution:

AB is the diameter.

‘O’ is centre of the circle, bisects AB

Let coordinates of A are (x, y)

As per the Mid-point formula,

∴ The coordinates of point A are (3, -10)

Question 8.

If A and B are (-2, -2) and (2, -4), respectively, find the coordinates of P such that AP = \(=\frac{3}{7}\) AB and P lies on the line segment AB.

Solution:

Let coordinates of P are (x, y).

AP = \(\frac{3}{7}\) AB AP : AB = 3 : 7

As per Section formula,

AP : AB = 3 : 7.

∴ AP + PB = AB

3 + PB = 7

∴ PB = = 7 – 3 = 4

∴ Let AP : PB = 3 : 4 = m_{1 }: m_{2}.

Coordinates of P are

Question 9.

Find the coordinates of the points which divide the line segment joining A(-2, 2) and B(2, 8) into four equal parts.

Solution:

Here, the given points are A(-2, 2) and B(2, 8)

Let P_{1}, P_{2} and P_{3} divide AB in four equal parts.

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Question 10.

Find the area of a rhombus if its vertices are (3, 0), (4, 5), (-1, 4) and (-2, -1) taken in order. [Hint: Area of a rhombus = \(\frac{1}{2}\) (product of its diagonals]

Solution:

ABCD is a rhombus.

AC and BD are the diagonals and intersects at ‘O’ and bisects perpendicularly.

∴ Let Diagonal BD = d_{1} Diagonal AC = d_{2},

We hope the given KSEEB SSLC Class 10 Maths Solutions Chapter 7 Coordinate Geometry Ex 7.2 will help you. If you have any query regarding Karnataka SSLC Class 10 Maths Solutions Chapter 7 Coordinate Geometry Exercise 7.2, drop a comment below and we will get back to you at the earliest