1. In what ratio is the segment joining the points (5, 1)and (−7, −1)divided by
x − axis?
(A) 1 ∶ 6
(B)6 ∶ 2
(C)2 ∶ 6
(D)1 ∶ 1
2. The coordinates of the points of trisection of a segment joining A(−3, 2)and
B(9, 5) is
(A)(3, 1), (−5, −4)
(B) (5, 9), (−9, 5)
(C)(2, 3), (4, 5)
(D) (1, 3), (5, 4)
3. A(−1, 2), B(4, 1)and C(7,6)are three vertices of the parallelogram ABCD.
The coordinates of fourth vertex is.
(A)(7, 2)
(B)(−2, 7)
(C)(7, −2)
(D) (2, 7)
4. If the distance between points (p, −7), (9, −7) is 15 units,then p is
(A) −3 or 7
(B) − 7 or 3
(C) − 3 or − 7
(D) − 6 or 24
5. Find the value of K if P(K,10)is the midpoint of AB, where the co − ordinate
of A and B are (−4, 13) and (8,7) respectively.
6. Find a relation between x and y such that the point (x ,y)is equidistant from the
points (7, 1) and (3, 5).
7. Prove that the points (−3, 0), (1, −3)and (4,1)are the vertices of an isosceles right
angled triangle. Find the area of this triangle.
8. Find the coordinates of the point which divides the line segment joining the points
(4, − 3) and (8, 5) in the ratio 3 ∶ 1 internally.
9. The line segment joining the points A(3,2)and B(5,1)is divided at the point P in the
ratio 1: 2 and P lies on the line 3x − 18y + k = 0. Find the value of k.
10. If (3,2)and (−3,2)are two vertices of an equilateral triangle which contains the
origin, find the third vertex.
Answers
Answer:
And 1 Given two points A(4,6) and (−7−1)
Let the coordinate of the point at X-axis be (x,o)
Let the ratio be m:n
By section formula:
m=6n ∴m:n=1:6
Ans 2 IDK
Ans 3 Let the three vertices of parallelogram ABCD be A(−1,2),B(4,1)andC(7,6)
Let the fourth vertex be D(a,b)
By joining AC and BD intersect at point O i.e, diagonals of parallelogram bisect each other
Mid point of AC=(3,4)
Mid point of BD=(3,4)
∴a=2,b=7
∴ The coordinate of fourth vertex =(2,7)
Ans 4 Use formula root (x2-x1)square + (y2-y1)square
Ans 5 IDK
Ans 6 Let P(x,y) be equidistant from the points A(7,1) and B(3,5).
AP=BP
AP^2=BP^2
(x−7) ^2 +(y−1)^2 =(x−3)^2 +(y−5)^2
x^2−14x+49+y^2−2y+1=x^2−6x+9+y^2
−10y+25
Hence x−y=2
Ans 7 IDK
Ans 8 We are given,
(x1,y1)=(4,−3) & 1(x2,y2)=(8,5)
Let (x,y)coordinates which divides the line joining the point (x1 ,y1 )and (x2 ,y2) in ratio m:n=3:1 internally.
So,(x,y)=(mx2+nx1/m+n , my2 +ny1/m+n)
=(3(8)+1(4)/3+1,3(5)+1(−3)/3+1)
=(28/4,12/4)
(x,y)=(7,3)
Ans 9 Given that point divides the line segment joining the points A(3,2) and B(5,1) in the ratio 1:2
The coordinate of P is ,
By section formula
( mx2+nx1/m+n, my2 +ny1/m+n )
here m:n =1:2,(x1q ,y1)=(3,2) and (x2,y2 )=(5,1)
∴ ( 1×5+2×3/1+2 , 1×1+2×2/1+2)=( 11/3, 5/3 )
3(11/3)−18(5/3)+K=0 K−19=0 ⇒ K=19
Ans 10
Let ΔABC be the equilateral triangle, where AB=BC=AC.
Vertices of this triangle be,
A(3,2),B(−3,2) and C(x,y).
The mid-point of the side AB is M(0,2).
AB= ✓(3+3)^2+(2−2)^2
AB=6 units
Therefore, AB=BC=AC=6 units
Also, AM=3 units
As two vertices are A(3,2) and B(−3,2), so third vertex will be at y−axis.
Therefore,
⇒C(0,y)
It will be located below the origin as the triangle contains the origin.
Now,
y^2 =(AC)^2 −(AM)^2
y^2 =36−9=27
y=3✓3units
Therefore,⇒C(0,−3 ✓3 )Hence, this is the required point.
Hope it helps you guys
Answer:
finding the same answer