Co­ordinate Geometry Exercise 14.1 (Part-14) | Extra Documents, Videos & Tests for Class 10 PDF Download

Question 27:Find the area of a parallelogram ABCD if three of its vertices are A(2, 4), B(2 + √33, 5) and C(2, 6).

Answer :It is given that A(2, 4), B(2 + √33, 5) and C(2, 6) are the vertices of the parallelogram ABCD.

We know that the diagonal of a parallelogram divides it into two triangles having equal area.
∴ Area of the parallogram ABCD = 2 × Area of the ∆ABC
Now,  

=√3square units 
∴ Area of the parallogram ABCD = 2 × Area of the ∆ABC = 2 × 
√3 = 2√3 square units
Hence, the area of given parallelogram is 2√3  square units

Question 28: Find the value(s) of k for which the points (3k − 1, k − 2), (k, k − 7) and (k − 1, −k − 2) are collinear.   

Answer : Let A(3k − 1, k − 2), B(k, k − 7) and C(k − 1, −k − 2) be the given points.
The given points are collinear. Then, 

ar(ΔABC)=0⇒1/2|x1(y2−y3)+x2(y3−y1)+x3(y1−y2)|=0⇒x1(y2−y3)+x2(y3−y1)+x3(y1−y2)=0

⇒(3k−1)[(k−7)−(−k−2)]+k[(−k−2)−(k−2)]+(k−1)[(k−2)−(k−7)]=0⇒(3k−1)(2k−5)+k(−2k)+5(k−1)=0⇒6k2−17k+5−2k2+5k−5=0⇒4k2−12k=0⇒3k-1k-7--k-2+k-k-2-k-2+k-1k-2-k-7=0⇒3k-12k-5+k-2k+5k-1=0⇒6k2-17k+5-2k2+5k-5=0⇒4k2-12k=0
⇒4k(k−3)=0⇒k=0 or k−3=0⇒k=0 or k=3⇒4kk-3=0⇒k=0 or k-3=0⇒k=0 or k=3Hence, the value of k is 0 or 3. 

Question 29: If the points A(−1, −4), B(b, c) and C(5, −1) are collinear and 2b + c = 4, find the values of b and c.

Answer : The given points A(−1, −4), B(b, c) and C(5, −1) are collinear.
∴ar(ΔABC)=0⇒12|x1(y2−y3)+x2(y3−y1)+x3(y1−y2)|=0⇒x1(y2−y3)+x2(y3−y1)+x3(y1−y2)=0∴ar∆ABC=0⇒12x1y2-y3+x2y3-y1+x3y1-y2=0⇒x1y2-y3+x2y3-y1+x3y1-y2=0⇒−1[c−(−1)]+b[−1−(−4)]+5(−4−c)=0⇒−c−1+3b−20−5c=0⇒3b−6c=21⇒b−2c=7               .....(1)⇒-1c--1+b-1--4+5-4-c=0⇒-c-1+3b-20-5c=0⇒3b-6c=21⇒b-2c=7               .....1Also, it is given that

2b + c = 4               .....(2)
Solving (1) and (2), we get
2(7+2c)+c=4⇒14+4c+c=4⇒5c=−10⇒c=−27+2c+c=4⇒14+4c+c=4⇒5c=-10⇒Putting c = −2 in (1), we get
b−2×(−2)=7⇒b=7−4=3b-2×-2=7⇒b=7-4=3Hence, the respective values of b and c are 3 and −2. 

Question 30: If the points A(−2, 1), B(a, b) and C(4, −1) ae collinear and a − b = 1, find the values of a and b.  

Answer : The given points A(−2, 1), B(a, b) and C(4, −1) are collinear.
∴ar(ΔABC)=0⇒12|x1(y2−y3)+x2(y3−y1)+x3(y1−y2)|=0⇒x1(y2−y3)+x2(y3−y1)+x3(y1−y2)=0∴ar∆ABC=0⇒12x1y2-y3+x2y3-y1+x3y1-y2=0⇒x1y2-y3+x2y3-y1+x3y1-y2=0⇒−2[b−(−1)]+a(−1−1)+4(1−b)=0⇒−2b−2−2a+4−4b=0⇒−2a−6b=−2⇒a+3b=1               .....(1)⇒-2b--1+a-1-1+41-b=0⇒-2b-2-2a+4-4b=0⇒-2a-6b=-2⇒a+3b=1               .....1Also, it is given that
a − b = 1               .....(2)
Solving (1) and (2), we get
b+1+3b=1⇒4b=0⇒b=0b+1+3b=1⇒4b=0⇒b=Putting b = 0 in (1), we get
a+3×0=1⇒a=1a+3×0=1⇒a=Hence, the respective values of a and b are 1 and 0. 

Question 31: Write the distance between the points A (10 cos θ, 0) and B (0, 10 sin θ).

Answer : 

We have to find the distance between A and B.

In general, the distance between A and B is given by,

So,

But according to the trigonometric identity,

Therefore,

Question 32: Write the perimeter of the triangle formed  by the points O (0, 0), A (a, 0) and B (0, b).

Answer : 

The distance d between two points  and  is given by the formula

The perimeter of a triangle is the sum of lengths of its sides.

The three vertices of the given triangle are O(0, 0), A(a, 0) and B(0, b).

Let us now find the lengths of the sides of the triangle.

The perimeter ‘P’ of the triangle is thus,

Thus the perimeter of the triangle with the given vertices is.

Question 35: If A (−1, 3) , B(1, −1) and C (5, 1) are the vertices of a triangle ABC, what is the length of the median through vertex A?

Answer : 

We have a triangle in which the co-ordinates of the vertices are A (−1, 3) B (1,−1) and

C (5, 1). In general to find the mid-point of two pointsand we use section formula as,

Therefore mid-point D of side BC can be written as,

Now equate the individual terms to get,

So co-ordinates of D is (3, 0)

So the length of median from A to the side BC,

Answer  : 

We have to find the unknown x using the distance between A and B which is 5.In general, the distance between A and B is given by,

So,

Squaring both the sides we get,

So,

Question 37: What is the area of the triangle formed by the points O (0, 0), A (6, 0) and B (0, 4)?

Answer : 

The given triangle is a right angled triangle, right angled at O. the co-ordinates of the vertices are O (0, 0) A (6, 0) and B (0, 4).

So,

Altitude is 6 units and base is 4 units.

Therefore,

Question 38: Write the coordinates of the point dividing line segment joining points (2, 3) and (3, 4) internally in the ratio 1 : 5.

Answer :  

Let P be the point which divide the line segment joining A (2, 3) and B (3, 4) in the ratio 1: 5.

Now according to the section formula if point a point P divides a line segment joining andin the ratio m: n internally than,

Now we will use section formula as,

So co-ordinate of P is

Question 39: If the centroid of the triangle formed by points P (a, b), Q(b, c) and R (c, a) is at the origin, what is the value of a + b + c?

Answer : 

The co-ordinates of the vertices are (a, b); (b, c) and (c, a)

The co-ordinate of the centroid is (0, 0)

We know that the co-ordinates of the centroid of a triangle whose vertices are  is-

So,

Compare individual terms on both the sides-

Therefore,

Question 40 : In Q. No. 9, what is the value of

Answer :  

The co-ordinates of the vertices are (a, b); (b, c) and (c, a)

The co-ordinate of the centroid is (0, 0)

We know that the co-ordinates of the centroid of a triangle whose vertices are  is-

So,

Compare individual terms on both the sides-

Therefore,

We have to find the value of -

Multiply and divide it by  to get,

Now as we know that if,

Then,

So,

Question 41:Write the coordinates of a point on X-axis which is equidistant from the points (−3, 4) and (2, 5).

Answer :

The distance d between two points  and  is given by the formula

Here we are to find out a point on the x−axis which is equidistant from both the points

A(-3,4) and B(2,5).

Let this point be denoted as C(x, y).

Since the point lies on the x-axis the value of its ordinate will be 0. Or in other words we have.

Now let us find out the distances from ‘A’ and ‘B’ to ‘C’

We know that both these distances are the same. So equating both these we get,

Squaring on both sides we have,

Hence the point on the x-axis which lies at equal distances from the mentioned points is.

Question 42: If the mid-point of the segment joining A (x, y + 1) and B (x + 1, y + 2) is C (3/2 , 5/2)32,52, find x, y.

Answer :

It is given that mid-point of line segment joining A and B is C

In general to find the mid-point of two pointsand we use section formula as,

So,

Now equate the components separately to get,

So,

Similarly,

So,

Question 43: Two vertices of a triangle have coordinates (−8, 7) and (9, 4) . If the centroid of the triangle is at the origin, what are the coordinates of the third vertex?

Answer : 

We have to find the co-ordinates of the third vertex of the given triangle. Let the co-ordinates of the third vertex be.

The co-ordinates of other two vertices are (−8, 7) and (9, 4)

The co-ordinate of the centroid is (0, 0)

We know that the co-ordinates of the centroid of a triangle whose vertices are  is-

So,

Compare individual terms on both the sides-

So,

Similarly,

So,

So the co-ordinate of third vertex

Question 44: Write the coordinates the reflections of points (3, 5) in X and Y -axes.

Answer :
We have to find the reflection of (3, 5) along x-axis and y-axis.Reflection of any pointalong x-axis isSo reflection of (3, 5) along x-axis isSimilarly, reflection of any pointalong y-axis isSo, reflection of (3, 5) along y-axis is

Question 45: If points Q and reflections of point P (−3, 4) in X and Y axes respectively, what is QR? 

Answer : 

We have to find the reflection of (−3, 4) along x-axis and y-axis.

Reflection of any pointalong x-axis is

So reflection of (−3, 4) along x-axis is

Similarly, reflection of any pointalong y-axis is

So, reflection of (−3, 4) along y-axis is

Therefore,

Question 46: Write the formula for the area of the triangle having its vertices at (x₁, y₁), (x2, y₂) and (x₃, y₃). 

Answer  : 

The formula for the area ‘A’ encompassed by three points,  and  is given by the formula,

The area ‘A’ encompassed by three points,  and  is also given by the formula,

Question 47: Write the condition of collinearity of points (x1, y1), (x2, y2) and (x3, y3).

Answer : 

The condition for co linearity of three points,  and  is that the area enclosed by them should be equal to 0.

The formula for the area ‘A’ encompassed by three points,  and  is given by the formula,

Thus for the three points to be collinear we need to have,

The area ‘A’ encompassed by three points,  and  is also given by the formula,

Thus for the three points to be collinear we can also have,

Question 48: Find the values of x for which the distance between the point P(2, −3), and Q (x, 5) is 10.

Answer :

It is given that distance between P (2,−3) and  is 10.

In general, the distance between A and B is given by,

So,

On further simplification,

Question 49: Write the ratio in which the line segment doining the points A (3, −6), and B (5, 3) is divided by X-axis.

Answer : 

Let P be the point of intersection of x-axis with the line segment joining A (3,−6) and B (5, 3) which divides the line segment AB in the ratio.

Now according to the section formula if point a point P divides a line segment joining andin the ratio m: n internally than,

Now we will use section formula as,

Now equate the y component on both the sides,

On further simplification,

So x-axis divides AB in the ratio 2:1.

Question 50:Find the distance between the points (−8/5 , 2)-85,2 and (2/5 ,2)25,2

Answer : 

We have to find the distance between and .

In general, the distance between A and B is given by,

So,