Class 10 Maths extra questions for Coordinate Geometry

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

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

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

(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

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

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)

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: XP²+ XQ² + XR²= CP²+CQ²+CR²+3CX²
L.H.S. = XP²+ XQ² + XR²
$(x � x_1)^2 + (y � y_1)^2 + (x � x_2)^2 + (y � y_2)^2 + (x � x_3)^2 + (y � y_3)^2$
$= 3x^2 + 3y^2 + x_1^2 + x_2^2 + x_3^2 + y_1^2 + y_2^2 + y_3^2 � 2x (x_1 + x_2 + x_3) � 2y(y_1 + y_2 + y_3)$
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. = CP²+CQ²+CR²+3CX²
$= (x_1 � 0)^2 + (y_1 � 0)^2 + (x_2 � 0)^2 + (y_2 � 0)^2 + (x_3 � 0)^2 + (y_3 � 0)^2 + 3[(x � 0)^2 + (y � 0)^2]$
$= x_1^2 + y_1^2 + x_2^2 + y_2^2 + x_3^2 + y_3^2 + 3(x^2 + y^2)$
$= 3x^2 + 3y^2 + x_1^2 + x_2^2 + x_3^2 + y_1^2 + y_2^2 + y_3^2$
= L.H.S.
Hence Proved

Mid points are given by
(-2, 3)=($ \frac {x_1 + x_2 }{2}$,$ \frac {y_1 + y_2 }{2}$ ) or $x_1 + x_2 = -4$ and $y_1 + y_2=6$
(4, -3)= ($ \frac {x_2 + x_3 }{2}$,$ \frac {y_2 + y_3 }{2}$ )or $x_2 + x_3 = 8$ and $y_2 + y_3=-6$
(4, 5)=($ \frac {x_1 + x_3 }{2}$,$ \frac {y_1 + y_3 }{2}$ )or $x_1 + x_3 = 9$ and $y_1 + y_3=10$
Coordinates of centroid G are ($ \frac {x_1 + x_2 + x_3}{3}$, $ \frac {y_1 + y_2 + y_3}{3}$ )
=($ \frac {2x_1 + 2x_2 + 2x_3}{6}$, $ \frac {2y_1 + 2y_2 + 2y_3}{6}$ )
=($ \frac {x_1 + x_2 + x_2+ x_3 + x_1 + x_3}{6}$, $ \frac {y_1 + y_2 + y_2+ y_3 + y_1 + y_3}{6}$ )
=($ \frac {-4 + 8 + 8}{6}$, $ \frac {6 -6 + 10}{6}$ )
=[ 2, 5/3]

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)

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)

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

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 (x₁, y₁), (x₂, y₂), (x₃, y₃) 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 x₁x₃x₃
$\frac {y_2 -y_3}{x_2x_3} + \frac {y_3 -y_1}{x_1x_3}+ \frac {y_1 -y_2}{x_2x_1}=0$

(1) C
(2) B
(3) D
(4) B