Coordinate Geometry Test

Result

Coordinate Geometry Test - 42

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

$$A(3,4)$$ and $$B(5,-2)$$ are two given points. If $$AP=PB$$ and area of $$\triangle {PAB}=10$$, then $$P$$ is -

A
$$(7,1)$$

B
$$(7,2)$$

C
$$(-7,2)$$

D
$$(-7,-1)$$

Solution

Let the coordinate P be $$(x,y)$$
Since it is given that $$PA=PB$$
So, by using distance formula
$$P(x,y)$$ and $$A(3,4)$$
$$PA=\sqrt{(3-x)^2+(4-y)^2}$$
$$P(x,y)$$ and $$B(5,-2)$$
$$PA=\sqrt{(5-x)^2+(-2-y)^2}$$
then,
$$\sqrt{(3-x)^2+(4-y)^2}=\sqrt{(5-x)^2+(-2-y)^2}$$
Squaring both sides
$$(3-x)^2+(4-y)^2=(5-x)^2+(-2-y)^2$$
$$9+x^2-6x+16+y^2-8y=25+x^2-10x+4+y^2+4y$$
$$-6x-8y=-10x+4+4y$$
$$4x-12y=4$$
$$x-3y=1.......(1)$$
Area of $$\triangle {PAB}=10$$
Area of triangle of (3,4),(5,-2) and (x,y)
Area of triangle $$=\dfrac{1}{2}[x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)]$$
$$10=\dfrac{1}{2}[3(-2-y)+5(y-4)+x(4+2)]$$
$$-6+2y-20+6x=20$$
$$6x+2y=46$$
$$3x+y=23......(2)$$
Since $$x-3y=1\Rightarrow x=3y+1$$
Equation (2) implies
$$3(3y+1)+y=23$$
$$9y+3+y=23$$
$$10y=20$$
$$y=2$$
$$x=3y+1$$
$$x=3\times2+1$$
$$x=7$$
Therefore, the coordinates are $$(7,2)$$.
Question 2
1 / -0

If the area of triangle formed by the points $$(2a,b),(a+b,2b+a)$$ and $$(2b,2a)$$ be $$\lambda$$ then the area of the triangle whose vertices are $$(a+b,a-b), (3b-a,b+3a)$$ and $$(3a-b,3b-a)$$ will be

A
$$\dfrac{3}{2} \lambda$$

B
$$3\lambda$$

C
$$4\lambda$$

D
$$none\ of\ these$$

Solution

Area of $$ \triangle \,= \dfrac{1}{2}\left | x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1})+x_{3}(y_{1}-y_{2}) \right |$$
Given area of triangle whose vertices are $$ (2a,b), (a+b,2b+a), (2b,2a)$$ is $$ \lambda $$
$$\therefore \lambda = \dfrac{1}{2}\left | 2a(2b+a-2a)+(a+b)(2a-b)+2b(b-2b-a) \right | $$
$$ \lambda = \dfrac{1}{2}\left | 4ab+2a^{2}-4a^{2}+2a^{2}-ab+2ab-b^{2}+2b^{2}-4b^{2}-2ab \right |$$
$$ \lambda = \dfrac{1}{2}\left | 3ab - 3b^{2} \right | $$ __ (1)
Now area of triangle whose vertices are $$ (a+b,a-b), (3b-a,b+3a), (3a-b,3b-a)$$ is given by
Area $$ = \dfrac{1}{2}\left | (a+b)(b+3a-3b+a)+(3b-a)(3b-a-a+b)+(3a-b)(a-b-b-3a) \right |$$
$$ = \dfrac{1}{2}\left | (a+b) (-2b+4a)+(3b-a)(4b-2a)+(3a-b)(-2a-2b) \right |$$
$$ = \dfrac{1}{2}\left | -2ab+4a^{2}-2b^{2}+4ab+12b^{2}-6ab-4ab+2a^{2}-6a^{2}-6ab+2ab+2b^{2} \right |$$
$$ = \dfrac{1}{2}\left | 12b^{2}-12ab \right |$$
$$ = \dfrac{4}{2}\left | -(3ab-3b^{2}) \right |$$
$$ = 4.\dfrac{1}{2}\left | 3ab-3b^{2} \right |$$
$$ = 4 \lambda .$$
Question 3
1 / -0

In a triangle $$ABC,A=(3,2),B=(1,4)$$ and the midpoint of $$AC$$ is $$(2,5)$$, then the mid point of $$BC$$ is

A
$$(3,4)$$

B
$$(1,8)$$

C
$$(1,6)$$

D
$$(3,6)$$

Solution

$$D(2,5)$$ is mid-point of AC
so using mid-point formula
$$\dfrac{x+3}{2}= 2$$
$$x+3=4$$
$$x=1$$
$$\dfrac{y+2}{2}=5$$
$$y+2=10$$
$$y=8$$
$$\therefore (x,y)= (1,8)$$
Let $$p(x_{1},y_{1})$$ be mid-point of $$BC$$
by using mid-point formula
$$x_{1}=\dfrac{1+1}{2}=\dfrac{2}{2}= 1$$
$$y_{1}=\dfrac{8+4}{2}= \dfrac{12}{2}=6$$
$$p(x_{1},y_{1})= (1,6)$$
Question 4
1 / -0

If $$A ( - 2, 5), B (3, 1)$$ and $$P,Q$$ are the points of intersection of $$AB$$ then mid-point of $$PQ$$ is

A
$$(2 , 3)$$

B
$$(\frac {1} {2} , 3)$$

C
$$(-\frac {1} {2} , 4)$$

D
$$(1 , 4)$$

Solution
Question 5
1 / -0

The area of the triangle vertices $$(1,0),(7,0)$$ and $$(4,4)$$ is ___ square units.

A
$$8$$

B
$$10$$

C
$$12$$

D
$$14$$

Solution
Question 6
1 / -0

If $$A = ( 3 , - 4 )$$ and the midpoints of $$A B , A C$$ are $$( 2 , - 1 ) , ( 4 , - 5 )$$ respectively then the mid.point of $$B C$$ is

A
$$( 1,2 )$$

B
$$( 3 , - 2 )$$

C
$$( - 1,2 )$$

D
$$( 0 , - 3 )$$

Solution

If $$P\left( {{x_1},{y_1}} \right)$$ and $$Q\left( {{x_2},{y_2}} \right)$$ are two points in the cartesian plane then their mid-point $$A\left( {{x},{y}} \right)$$ is given by,
$$\left( {x , y} \right) = \left( {\dfrac{{{x_1} + {x_2}}}{2},\dfrac{{{y_1} + {y_2}}}{2}} \right)$$

Given:
Mid point of $$AB=(2, -1)$$
$$A=(3, -4)$$
Let the coordinates of the point $$B$$ be $$(x_2, y_2).$$
$$\left(\dfrac{3+x_2}{2}, \dfrac{y_2-4}{2}\right)=(2, -1)$$
$$\Rightarrow x_2+3=4$$ and $$y_2-4=-2$$
$$\Rightarrow x_2=1$$ and $$y_2=2$$
$$\therefore B(1, 2)$$

Also,
$$A=(3, -4)$$
Mid point of $$AC$$ $$=(4, -5)$$
Let the coordinates of the point $$C$$ be$$(x_3, y_3).$$
$$\left(\dfrac{3+x_3}{2}, \dfrac{y_3-4}{2}\right)=(4, -5)$$
$$\Rightarrow x_3 + 3=8$$ and $$y_3 - 4=-10$$
$$\Rightarrow x_3 = 5$$ and $$y_3 = -6$$
$$\therefore C(5, -6)$$

Now, by using the coordinates of the point $$B$$ and $$C,$$
Mid point of $$BC$$ $$=\left(\dfrac{1+5}{2}, \dfrac{2-6}{2}\right)$$
$$=\left(\dfrac{6}{2}, \dfrac{-4}{2}\right)$$
$$=(3, -2)$$
Question 7
1 / -0

The co -ordinates of the midpoint of a line segment joining $$ P(5,7) $$ and $$ Q (-3,3) $$ are........

A
$$ (2,4) $$

B
$$ (1,5 ) $$

C
$$ (4,2 ) $$

D
$$ (2,5 ) $$

Solution

Given,

$$P(5,7),Q(-3,3)$$

mid point is given by,

$$(x,y)=\left ( \dfrac{5-3}{2},\dfrac{7+3}{2} \right )$$

$$\Rightarrow (x,y)=(1,5)$$
Question 8
1 / -0

The area of the quadrilateral formed by the points $$(1,1),(7,-3),(12,2)$$ and $$(7,21)$$ is

A
$$130$$

B
$$132$$

C
$$\dfrac{131}{2}$$

D
$$264$$

Solution
Question 9
1 / -0

The point which divides the line segment joining $$(-2, 4), (2, 7)$$ in the ratio $$2:1$$ internally is:

A
$$(6, 10)$$

B
$$(\dfrac{2}{3}, \dfrac{14}{3})$$

C
$$(\dfrac{-4}{3}, \dfrac{2}{3})$$

D
$$( \dfrac{2}{3} ,6)$$

Solution
Question 10
1 / -0

In the given figure (not drawn to scale), the coordinates of P, Q and R are (3, 0) (0, 5) and (0, -5) respectively. Find the area of $$\Delta PQR$$.

A
20 sq. units

B
40 sq. units

C
15 sq. units

D
25/2 sq. units

Solution