Math, asked by madfangirl, 7 months ago

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

Answered by nikkijonas80
6

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

Answered by tanujpatel085
0

Answer:

finding the same answer

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