Ncert – Class 10 –Coordinate Geometry-Hots Question

Question 1

Seema is the student of a prestigious school of her city. One day her school decided to take the students in an educational trip in near by sugarcane field. One farmer give her a sugarcane when they are returning to their school. Seema decided to equally distribute this sugarcane between her mother, father, younger sister and herself. If the end points of sugarcane are A(3,2) and B (7,5) , then find the coordinates of the points which divide the sugarcane into four equal parts.

Solution

Mid point( Point C) of AB= (5,3.5)
Mid point of AC= (4,2.75)
Mid point of BC= (6,4.25)

Question 2

In a coordinate system, P = (2, 7) and Q = (2, –3). Which could be the coordinates of R if PQR is an isosceles triangle?
I. (12, –3)
II. (–6, –9)
III. (–117, 2)

Solution

I, II, and III

Question 3

Show that the four points (0, -1), (6, 7), (-2, 3) and (8, 3) are the vertices of a rectangle.

Solution

Let A(0,-1),B(6,7),C(-2,3) and D(8,3) be the given points .
Then by distance formula
AD=4√5 unit
BC=4√5 unit
AC=2√5 unit
BD=2√5 unit
Here AD=BC and AC=BD
So ,ADBC is a parallelogram,
AB=10 unit
CD=10 unit
Clearly , (𝐴𝐵)²=(𝐴𝐷)²+ (𝐷𝐵)²
(𝐶𝐷)²=(𝐶𝐵)²+(𝐵𝐷)²
Hence ADBC is a rectangle.

Question 4

If the two vertices of an equilateral triangle be (0, 0), (3,√3) then find the third vertex.

Solution

O(0,0) and A(3,√3) be the given points and let B(x,y) be the 3
third vertex of equilateral triangle OAB.
Then OA=OB=AB
Or,(𝑂𝐴)²=(𝑂𝐵)²=(𝐴𝐵)²
(𝑂𝐴)² = 12 , (𝑂𝐵)² =𝑥²+𝑦²

(𝐴𝐵)²=x²+y²-6x-2√3y+12
(𝑂𝐴)²=(𝑂𝐵)² and(𝑂𝐵)² =(𝐴𝐵)²
i.e. x²+y²=12 and x²+y²=x²+y² -6x-2√3y +12
Or, x²+y²=12 and 6x+2√3y=12
Now, x²+((6-3x)/2)²=12 since y=(6-3x)/2
Then, x=0,3
Putting x in y=(6-3x)/2
Y=2√3 , √3

Question 5

Let the opposite angular points of a square be (3, 4) and (1, -1). Find the coordinates of the remaining angular points

Solution

[(23-10Y)/4]² + Y² – (23- 10Y) – 3Y – 1= 0
4Y – 12Y +5 = 0
(2Y -1)(2Y-5) = 0
–>Y = ½ OR 5/2
Putting Y = ½ and Y = 5/2 respectively in (1)
we get X = 9/2 and X= -1/2
hence the required vertices of the squares are
( 9/2, 1/2)and (-1/2 , 5/2)

Question 6

If A(-2,1),B(a,0), C(4, b) and D(1,2) are the vertices of a parallelogram ABCD. Find the values of a and b. Hence find the length of its sides

Solution

a=1, b=1,length= √10

Question 7

In what ratio does the point P(-4,y) divide the line segment joining the point A (-6, 10) and B(3,-8) .Hence find the value of y

Solution

2:7, y =6

Question 8

Prove that (a,a), (-a,-a) and (-√3a, √3a) are the vertices of an equilateral triangle.

Solution

For correct proof proportionate mark

Question 9

Point A lies on the line segment PQ joining P(6, -6) and Q(-4, -1) in such a way that PA/PQ=2/5. If point P also lies on the line 3x + k(y + 1) = 0, find the value of k

Solution

Coordinates of P are (6,-6) given that
Since, (6,-6) lies on the line so,
3x + k (y+1) =0
3 x 6 + k (-6 + 1) =0
18 – 5k +0
18 = 5k
K=18/5

Question 10

If the point P(x, y) is equidistant from the points A(a + b, b – a) and B(a – b, a + b), prove that bx = ay

Solution

PA = PB … [Given
PA2 = PB2 … [Squaring both sides
⇒ [(a + b) – x]² + [(b a) – y)² = [(a – b) – x]²+ [(a + b) – y]²
⇒ (a + b)² + x² – 2(a + b)x + (b – a)² + y2 – 2(b – a)y = (a – b)² + x² – 2(a – b)x + (a-b)² + y² – 2(a + b)y …

[∵ (a – b) ² = (b – a)²
⇒ -2(a + b)x + 2(a – b)x = -2(a + b)y + 2(b – a)y
⇒ 2x(-a – b + a – b) = 2y(-a – b + b – a)
⇒ -2bx = – 2ay
⇒ bx = ay (Hence proved)