Coordinate Geometry NCERT Exercise 7.2 part two Class Ten Mathematics

Coordinate Geometry

Exercise 7.2 Part 2

Question 5: Find the ratio in which 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: We have; `x_1 = 1`, `y_1 = 5`, `x_2 = -4`, `y_2 = 5`

Coordinates of a point on x-axis = (x, 0)

Let us assume the ratio = k : 1

`x=(1-4k)/(k+1)`-------(1)

`0=(-5+5k)/(k+1)`

Or, `-5+5k=0`
Or, `5k=5`
Or, `k=1`

Substituting this value in equation (1), we get;

`x=(1-4)/(1+1)=-3/2`

Ratio = (1 : 1)

Point C = (-3/2, 0)


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: A = (1, 2), B = (4, y), C = (x, 6), D = (3, 5)

AC = BD (diagonals of parallelogram are equal)

We know that diagonals of a parallelogram bisect each other. Let us assume point O is the point of intersection of two diagonals. Then coordinates of point O can be given as follows:

For diagonal AC;

`O=((1+x)/(2)\, (2+6)/(2))`

`=((1+x)/(2)\, 4)`--------(1)

For diagonal BD;

`O=((3+4)/(2)\, (5+y)/(2))`

`=((7)/(2)\, (5+y)/(2))`-------(2)

From equations (1) and 2;

`(1+x)/(2)=7/2`

Or, `1+x=7`
Or, `x=6`
`(5+y)/(2)=4`

Or, `5+y=8`
Or, `y=3`

Hence, x = 6, y = 3


Question 7: Find the coordinates of a point A, where AB is the diameter of a circle, whose centre is (2, -3) and B is (1, 4).

Solution: Centre O bisects AB and AB = (x, y). Hence;

`2=(x+1)/(2)`

Or, `x+1=4`
Or, `x=3`

Similarly,

`-3=(y+4)/(2)`

Or, `y+4=-6`
Or, `y=-10`

Hence; O = (3, -10)

Question 8: If A and B are (-2, -2) and (2, -4) respectively, find the coordinates of P such that AP = 3/7(AB) and P lies on the line segment AB.

Solution:

`AP=3/7\AB`

Or, `(AP)/(AB)=3/7`

Or, `(AP)/(PB)=3/4`

This means that point P divides AB in 3 : 4 ratio. Coordinates of P can be calculated as follows:

`x=(4xx(-2)+3xx2)/(7)`

`=(-8+6)/(7)=-2/7`

`y=(4xx(-2)+3xx(-4))/(7)`

`=(-8-12)/(7)=-20/7`

Hence, P = (-2/7, -20/7)


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.

10 coordinate exercise solution

Solution: Let us assume point Q bisects AB; as shown in figure.

Coordinates of Q can be calculated as follows:

`x=(2-2)/(2)=0`

`y=(2+8)/(2)=5`

Or, `Q=(0,5)`

Now, point P bisects AQ.

Coordinates of P can be calculated as follows:

`x=(-2+0)/(2)=-1`

`y=(2+5)/(2)=7/2`

Or, `P=(-1,\7/2)`

Point R bisects QB.

Coordinates of R can be calculated as follows:

`x=(0+2)/2=1`

`y=(5+8)/(2)=13/2`

Or, `R(1,\13/2)`

Hence, P = (-1, 7/2), Q = (0, 5) and R = (1, 13/2)

Question 10: Find the area of a rhombus if its vertices are (3, 0), (4, 5), (-1, 4) and (-2, -1) taken in order.

Solution: A = (3, 0), B = (4, 5), C = (-1, 4), D = (-2, -1)

Diagonal AC can be calculated as follows by using distance formula:

`AC=sqrt((x_2-x_1)^2+(y_2-y_1)^2)`

`=sqrt((-1-3)^2+(4-0)^2)`

`=sqrt((-4)^2+4^2)`

`=sqrt(16+16)=4sqrt2`

Similarly, diagonal BD can be calculated as follows:

`BD=sqrt((x_2-x_1)^2+(y_2-y_1)^2)`

`=sqrt((-2-4)^2+(-1-5)^2)`

`=sqrt((-6)^2+(-6)^2)`

`=sqrt(36+36)=6sqrt2`

Area of rhombus can now be calculated as follows:

`1/2\xxd_1xxd_2`

`=1/2\xx4sqrt2xx6sqrt2=24 sq` unit



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