Class 10 Maths extra questions for Coordinate Geometry
Given below are the Class 10 Maths extra questions and Important Questions for Coordinate Geometry
(a)Short answer type
(b)True or False statement
(c) Long answer questions
(d) Multiple Choice Questions
Short answer type
Question 1
Calculate the Following
Distance between the point (1,3) and ( 2,4)
Mid-point of line segment AB where A(2,5) and B( -5,5)
Area of the triangle formed by joining the line segments (0,0) ,( 2,0) and (3,0)
Distance of point (5,0) from Origin
Distance of point (5,-5) from Origin
Coordinate of the point M which divided the line segment A(2,3) and B( 5,6) in the ratio 2:3
Quadrant of the Mid-point of the line segment A(2,3) and B( 5,6)
the coordinates of a point A, where AB is the diameter of circle whose center is (2,−3) and B is (1, 4)
Solution
(a) $D=\sqrt{(1-2)^{2}+(3-4)^{2}}=\sqrt{2}$
(b)Mid-point is given by (2-5)/2,(5+5)/2 or (-3/2, 5)
(c) $A=\frac{1}{2}[0(0-0)+2(0-0)+3(0-0)]=0$
Since the three points are collinear, the area is zero
(d) $D=\sqrt{5^{2}+0^{2}}=5$
(e)$D=\sqrt{(-5)^{2}+0^{2}}=5$
(f) Coordinates of point M is given by
$x=\frac{2X3+3X2}{2+3}=\frac{12}{5}$
$y=\frac{2X6+3X5}{2+3}=\frac{27}{5}$
(g) Mid point is given by (7/2, 9/2) which lies in First quadrant
(h) We know that center is mid point of AB, So
$2=\frac{1+x}{2}$
$-3=\frac{4+y}{2}$
Solving these, we get (3,-10)
True or False statement
Question 2
(a) Point A( 0,0) B( 0,3) ,C( 0,7) and D( 2,0) formed a quadrilateral
(b) The point P (-2, 4) lies on a circle of radius 6 and center C (3, 5)
(c) Triangle PQR with vertices P (-2, 0), Q (2, 0) and R (0, 2) is similar to Δ XYZ with
Vertices X (-4, 0) Y (4, 0) and Z (0, 4).
(d) Point X (2, 2) Y (0, 0) and Z (3, 0) are not collinear
(e) The triangle formed by joining the point A( -3,0) , B( 0,0) and C( 0,2) is a right angle triangle
(f) A circle has its center at the origin and a point A (5, 0) lies on it. The point B (6, 8) lies inside the circle
(g) The points A (-1, -2), B (4, 3), C (2, 5) and D (-3, 0) in that order form a rectangle Solution
False, As three point are A,B and C are collinear, So no quadrilateral can be formed
False, As the distance between the point P and C is $\sqrt{26}$ which is less than 6.So point lies inside the circle
True. Both the triangle are equilateral triangle with side 4 and 8 respectively
True. As the Area formed by the triangle XYZ is not zero
True, If we plot the point on the Coordinate system, it becomes clear that it is right angle at origin
False. The radius of the circle is 5 and distance of the point B is more than 5,So it lies outside the circle
True. If we calculate the distance between two points, it becomes clear that opposite side are equal, also the diagonal are equal. So it is a rectangle
Long Answer Type
Question 1
Determine the ratio in which the point P (m, 6) divides the join of A (-4, 3) and B (2, 8). Also find the value of m. Solution
Let point P divides AB in ratio k:1
Then by section formula
$x= \frac {mx_2+nx_1}{m+n}$
or
$m= \frac {2k -4}{k+1}$
Also
$y= \frac {my_2+ny_1}{m+n}$
?$6= \frac {8k+3}{k+1}$
or
k=3/2
Therefore value of m will be
$m= \frac {2(3/2) -4}{(3/2)+1} = \frac {-2}{5}$
Ratio, k:1 = 3/2 : 1= 3 :2
Question 2
The line segment joining the points (3, -4) and (1, 2) is trisected at the points P and Q. If the co-ordinates of P and Q are (p, -2) and (5/3, q) respectively. Find the values of p and q. Solution
As the line segment is trisected at the points P and Q, P will divided the point in 1 :2 and Q in 2: 1
So Applying section Formula,
Coordinates of P (x, y) = ( $\frac {mx_2+nx_1}{m+n}$,$\frac {my_2+ny_1}{m+n}$)
=($\frac {1 \times 1 + 2 \times 3}{1+2}$,$\frac {1 \times 2 + 2 \times -4}{1+2}$
=(7/3 ,-2)
Now P points are given as (p, -2), So comparing p =7/3
Similarly Coordinates of Q (x, y) = ( $\frac {mx_2+nx_1}{m+n}$,$\frac {my_2+ny_1}{m+n}$)
==($\frac {2 \times 1 + 1 \times 3}{2+1}$,$\frac {2 \times 2 + 1 \times -4}{2+1}$
=(5/3 ,0)
Now Q points are given as (5/3, q), So comparing q =0
Question 3
The line joining the points (2, 1) and (5, -8) is trisected at the points P and Q. If point P lies on the line 2x - y + K = 0. Find the value of K. Solution
As the line segment is trisected at the points P and Q, P will divided the point in 1 :2 and Q in 2: 1
So Applying section Formula,
Coordinates of P (x, y) = ( $\frac {mx_2+nx_1}{m+n}$,$\frac {my_2+ny_1}{m+n}$)
=($\frac {1 \times 5 + 2 \times 2}{1+2}$,$\frac {1 \times -8 + 2 \times 1}{1+2}$
=(3,-2)
Now if P lies on line x - y + K = 0, then
6 + 2 + k=0
k =-8
Question 4
A (4, 2), B (6, 5) and C (1, 4) are the vertices of ΔABC.
(i)The median from A meets BC in D. Find the co- ordinates of the point D.
(ii)Find the co- ordinates of point P on AD such that AD : PD = 2 : 1
(iii)Find the co- ordinates of the points Q and R on medians BE and CF respectively such that BQ : QE = 2 : 1 and CR : RF = 2 : 1
(iv) What do you observe? Solution
(i) Median is defined as the line joining the midpoint of one side of a triangle to the opposite vertex
So coordinates of D will be the mid point of B (6, 5) and C (1, 4)
=($ \frac {6+1}{2}$, $\frac {5+4}{2}$)
=[7/2,9/2 ]
(ii) Coordinates of point P will be given by the section Formula
=( $\frac {mx_2+nx_1}{m+n}$,$\frac {my_2+ny_1}{m+n}$)
=($\frac {2 \times (7/2)+1 \times 4}{2+1}$,$\frac {2 \times (9/2)+ 1 \times 2}{2+1}$)
=[11/3,11/3]
(iii) coordinates of E will be the mid point of A (4, 2) and C (1, 4)
=($ \frac {4+1}{2}$, $\frac {2+4}{2}$)
=[5/2,3 ]
coordinates of F will be the mid point of A (4, 2) and B (6, 5)
=($ \frac {4+6}{2}$, $\frac {2+5}{2}$)
=[5,7/2 ]
Coordinates of point Q will be given by the section Formula
=( $\frac {mx_2+nx_1}{m+n}$,$\frac {my_2+ny_1}{m+n}$)
=[11/3,11/3]
Coordinates of point R will be given by the section Formula
=( $\frac {mx_2+nx_1}{m+n}$,$\frac {my_2+ny_1}{m+n}$)
=[11/3,11/3]
(iv) The coordinates of Points P,Q,R are same [11/3,11/3].
This is called the centroid of the triangle
Question 5
If the points A (6, 1), B (8, 2), C (9, 4) and D (k, p) are the vertices of a parallelogram taken in order, then find the value of k and p. Solution
In a parallelogram, diagonal bisect each other.
Mid point of A (6, 1) and C (9, 4)
=($ \frac {6+9}{2}$, $\frac {1+4}{2}$)
=[15/2,5/2 ]
Mid point of B (8, 2) and D (k, p)
=($ \frac {8+k}{2}$, $\frac {2+p}{2}$)
Now both should be same
Therefore
$\frac {8+k}{2}= \frac {15}{2}$
k=7
$\frac {2+p}{2}=\frac {5}{2}$
p=3
Question 6
Two vertices of a triangle are (1, 2), (3, 5) and its centroid is at the origin. Find the co-ordinates of the third vertex. Solution
coordinates of centroid G are ($ \frac {x_1 + x_2 + x_3}{3}$, $ \frac {y_1 + y_2 + y_3}{3}$ )
Therefore
$0=1 + 3 + x_3$ or $x_3=-4$
$0=2 + 5 + y_3$ or $y_3=-7$
Coordinates of Third vextex is (-4,-7)
Question 7
If C be the centroid of a triangle PQR and X be any other point in the plane, prove that
XP^{2}+ XQ^{2} + XR^{2}= CP^{2}+CQ^{2}+CR^{2}+3CX^{2} Solution
Let the coordinates of A, B and C be ($x_1$,$y_1$), ($x_2$,$y_2$) and ($x_3$,$y_3$)
Then coordinates of centroid G will be ($ \frac {x_1 + x_2 + x_3}{3}$, $ \frac {y_1 + y_2 + y_3}{3}$ )
Let us the assume that centroid lies on the origin
Therefore
$x_1 + x_2 + x_3=0$,$y_1 + y_2 + y_3=0$
Let the coordinates of point X be (x, y) To Proof: XP^{2}+ XQ^{2} + XR^{2}= CP^{2}+CQ^{2}+CR^{2}+3CX^{2}
L.H.S. = XP^{2}+ XQ^{2} + XR^{2}
$(x � x_1)^2 + (y � y_1)^2 + (x � x_2)^2 + (y � y_2)^2 + (x � x_3)^2 + (y � y_3)^2$
$= 3x^2 + 3y^2 + x_1^2 + x_2^2 + x_3^2 + y_1^2 + y_2^2 + y_3^2 � 2x (x_1 + x_2 + x_3) � 2y(y_1 + y_2 + y_3)$
Now as $x_1 + x_2 + x_3=0$,$y_1 + y_2 + y_3=0$
$= 3x^2 + 3y^2 + x_1^2 + x_2^2 + x_3^2 + y_1^2 + y_2^2 + y_3^2$
R.H.S. = CP^{2}+CQ^{2}+CR^{2}+3CX^{2}
$= (x_1 � 0)^2 + (y_1 � 0)^2 + (x_2 � 0)^2 + (y_2 � 0)^2 + (x_3 � 0)^2 + (y_3 � 0)^2 + 3[(x � 0)^2 + (y � 0)^2]$
$= x_1^2 + y_1^2 + x_2^2 + y_2^2 + x_3^2 + y_3^2 + 3(x^2 + y^2)$
$= 3x^2 + 3y^2 + x_1^2 + x_2^2 + x_3^2 + y_1^2 + y_2^2 + y_3^2$
= L.H.S.
Hence Proved
Question 8
If P(x, y) is any point on the line joining the points A (a, 0), B (0, b), then show that x/a + y/b = -1
Question 9
If (-2, 3) (4, -3) and (4, 5) are the mid points of the sides of a triangle, find the co-ordinates of its centroid. Solution
Question 10
Find the third vertex of a triangle, if two of its vertices are at (-3, 1) and (0, -2) and the centroid is at the origin. Solution
coordinates of centroid G are ($ \frac {x_1 + x_2 + x_3}{3}$, $ \frac {y_1 + y_2 + y_3}{3}$ )
Therefore
$0=0 - 3 + x_3$ or $x_3=3$
$0=1 -2 + y_3$ or $y_3=1$
Coordinates of Third vextex is (3,1)
Question 11
A (3, 2) and B (-2, 1) are two vertices of a triangle ABC whose centroid G has the co- ordinates [5/3,-1/3]. Find the co- ordinates of the third vertex C of the triangle. Solution
coordinates of centroid G are ($ \frac {x_1 + x_2 + x_3}{3}$, $ \frac {y_1 + y_2 + y_3}{3}$ )
Therefore
$\frac {5}{3}=\frac {3 - 2 + x_3}{3}$ or $x_3=4$
$\frac {-1}{3}=\frac {2 +1 + y_3}{3}$ or $y_3=-4$
Coordinates of Third vextex is (4,-4)
Question 12
If D, E and F are the mid points of sides BC, CA and AB respectively of ΔABC, then testing co- ordinate geometry prove that
Area of ΔDEF =(1/4) &(Area of ΔABC) Question 13
If A (4, 6), B (3, -2) and C (5, 2) are the vertices of ΔABC, then verify the fact that a median of a triangle ABC divides it into two triangles of equal areas. Question 14
The area of a triangle is 5. Two of its vertices are (2, 1) and (3, 2). The third vertex lies on y = x + 3. Find the third vertex. Question 15
Four points A (6, 3), B (-3, 5), C (4, -2) and D (x, 3x) are given in such a way that
$\frac {\Delta DBC}{\Delta ABC} = \frac {1}{2}$
find x. Solution
Area of triangle is given by
$A= \frac {1}{2} [x_1(y_2 - y_3) + x_2(y_3 -y_1) + x_3(y_1 - y_2)]$
Area of triangle Δ DBC
$A_1= \frac {1}{2} [x(5 +2) -3(-2 -3x) + 4(3x - 5)]$
$=[14x -7]$
Area of triangle Δ ABC
$A_2= \frac {1}{2} [6(5 +2) -3(-2 -3) + 4(3 - 5)]$
$=\frac {49}{2}$
Now As per the question
$\frac {14x-7}{49/2} = \frac {1}{2}$
$4(14x -7) =49$
$4(2x -1) =7$
$8x -4=7$
x=11/8
Question 16
If three points (x_{1}, y_{1}), (x_{2}, y_{2}), (x_{3}, y_{3}) lie on the same line, prove that
$\frac {y_2 -y_3}{x_2x_3} + \frac {y_3 -y_1}{x_1x_3}+ \frac {y_1 -y_2}{x_2x_1}=0$ Solution
Area of triangle is given by
$A= \frac {1}{2} [x_1(y_2 - y_3) + x_2(y_3 -y_1) + x_3(y_1 - y_2)]$
If three points (x_{1}, y_{1}), (x_{2}, y_{2}), (x_{3}, y_{3}) lie on the same line,then
A=0
$\frac {1}{2} [x_1(y_2 - y_3) + x_2(y_3 -y_1) + x_3(y_1 - y_2)]=0$
Dividing by x_{1}x_{3}x_{3}
$\frac {y_2 -y_3}{x_2x_3} + \frac {y_3 -y_1}{x_1x_3}+ \frac {y_1 -y_2}{x_2x_1}=0$
Question 17
Ifthe point P (m, 3) lies on the line segment joining the points A [-2/5, 6] and B (2, 8), find the value of m. Question 18
If R (x, y) is a point on the line segment joining the points P (a, b) and Q (b, a) then prove that x + y = a + b Question 19
Find the value of a for which the area of the triangle formed by the points A (a, 2a), B (-2, 6) and C (3, 1) is 10 square units. Question 20
If G be the centroid of a triangle ABC, prove that
AB^{2} + BC^{2} + CA^{2} = 3(GA^{2}+ GB^{2}+ GC^{2}) Question 21
Find the co-ordinates of circumcenter of a Δ ABC where A( 1, 2) ,B ( 3, -4) and C ( 5, -6 ) Question 22
The coordinates of the mid-point of the line joining the points (2p + 2, 3) and (4, 2q + 1) are (2p, 2q). Find the value of p and q.
Multiple Choice Questions
Question 1
The area of a triangle with vertices A (3, 0), B (7, 0) and C (8, 4) is
(a) 14
(b) 28
(c) 8
(d) 6 Question 2
The points (�4, 0), (4, 0), (0, 3) are the vertices of a
(a) right triangle
(b) isosceles triangle
(c) equilateral triangle
(d) scalene triangle Question 3
The point which divides the line segment joining the points (7, �6) and (3, 4) in ratio 1 : 2 internally lies in the
(a) I quadrant
(b) II quadrant
(c) III quadrant
(d) IV quadrant Question 4
If the distance between the points (2, �2) and (�1, x) is 5, one of the values of x is
(a) �2
(b) 2
(c) �1
(d) 1 Solution 1-4
(1) C
(2) B
(3) D
(4) B
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