Free Coordinate Geometry Subjective Test 02 Practice Test - 10th Grade 

Concept : 1 Mark
Application : 1 Mark

Let the coordinates of the third vertex be (x, y). 

We know that the coordinates of the centroid of a triangle whose vertices are

$(x_1,y_1)$,$(x_2,y_2)$ and $(x_3,y_3)$ are

$\frac{x_1+x_2+x_3}{3},\frac{y_1+y_2+y_3}{3}$

Given points are (3, –5) and (–7, 4) and centroid is (2, –1).

$\Rightarrow \frac{x+3-7}{3}=2$ and $\frac{y-5+4}{3}=-1$

$\Rightarrow x-4=6$ and $y-1=-3$

$\Rightarrow x=10$ and $y=-2$

Thus the coordinates of the third vertex are (10, -2).

Concept : 1 Mark
Application : 1 Mark

Distance between points $A(x_1,y_1)$ and $B(x_2,y_2)$ is,

$D=\sqrt{{(x_2-x_1)}^2+{(y_2-y_1)}^2}$

$D=\sqrt{{(4-(-3\left)\right)}^2+{(5-(-3\left)\right)}^2}$

$D=\sqrt{7^2+8^2}$

$D=\sqrt{113}units$

Concept : 1 Mark
Application : 1 Mark

Let p (x , -1) and Q (3 ,2) be the given points. Then,

PQ = 5 (Given)
Distance between the points is given by
$\sqrt{{(x_1-x_2)}^2+{(y_1-y_2)}^2}$
$\Rightarrow \sqrt{{(x-3)}^2+{(-1-2)}^2}=5$

$\Rightarrow {(x-3)}^2+9=5^2$     [Squaring both sides]

$\Rightarrow x^2-6x+18=25$ 

$\Rightarrow x^2-6x-7=0$

$\Rightarrow (x-7)(x+1)=0$

$\Rightarrow x=7orx=-1$

Concept : 1 Mark
Application : 1 Mark

A (1 , -1) , B (5 , 2) and C (9 , 5) are the given points. Then, We have 
Distance between the points is given by
$\sqrt{{(x_1-x_2)}^2+{(y_1-y_2)}^2}$
$AB=\sqrt{{(5-1)}^2+{(2+1)}^2}$
So,

$AB=\sqrt{16+9}=5$

$BC=\sqrt{{(5-9)}^2+{(2-5)}^2}$

$BC=\sqrt{16+9}=5$

and , $AC=\sqrt{{(1-9)}^2+{(-1-5)}^2}$

$AC=\sqrt{64+36}=10$

Clearly, AC = AB + BC

Hence A, B, C are collinear points

Formula: 1 Mark
Concept:1 Mark
Answer: 1 Mark

Let the coordinates of B be $(\alpha ,\beta )$.

It is given that AC:BC = 3:4.

So, the coordinates of C are:

$C(x,y)=\left(\frac{m{x}_2+n{x}_1}{m+n}\right),\left(\frac{m{y}_2+n{y}_1}{m+n}\right)$

$\Rightarrow C(-1,2)=\frac{3\alpha +4\times 2}{3+4},\frac{3\beta +4\times 5}{3+4}$

$\Rightarrow C(-1,2)=\frac{3\alpha +8}{7},\frac{3\beta +20}{7}$

$\Rightarrow \frac{3\alpha +8}{7}=-1and\frac{3\beta +20}{7}=2$

$\Rightarrow \alpha =-5and\beta =-2$

Thus, the coordinates of B are (-5, -2).

Concept: 1 Mark
Application: 1 Mark
Answer: 1 Mark 

Let the required ratio be $\lambda :1$. Then, the coordinates of the point of division are,

$R\left(\frac{5\lambda +2}{\lambda +1},\frac{6\lambda -3}{\lambda +1}\right)$

But, it is a point on x-axis on which y-coordinates of every point is zero.

$\therefore \frac{6\lambda -3}{\lambda +1}=0\Rightarrow \lambda =\frac{1}{2}$

Thus, the required ratio is $\frac{1}{2}:1$ or 1:2.

Substituting $\lambda =\frac{1}{2}$ in the coordinates of R.

$R\left(\frac{5\lambda +2}{\lambda +1},0\right)$

$\Rightarrow R\left(\frac{\frac{5}{2}+2}{\frac{1}{2}+1},0\right)$

$\Rightarrow R(\frac{\frac{9}{2}}{\frac{3}{2}},0)$

$\Rightarrow R(3,0)$

$\therefore$ the coordinates of R are (3, 0)

Formula: 1 Mark
Application: 1 Mark
Answers: 1 Mark 

We know that the diagonals of a parallelogram bisect each other.

$\therefore$ The coordinates of the mid-point of AC = The coordinates of the mid-point of BD i.e.

If $A(x_1,y_1),B(x_2,y_2),C(x_3,y_3)$ and $D(x_4,y_4)$ are the vertices of parallelogram ABCD, Then

$\frac{x_1+x_3}{2},\frac{y_1+y_3}{2}=\frac{x_2+x_4}{2},\frac{y_2+y_4}{2}$

$\Rightarrow (\frac{-2+4}{2},\frac{-1+b}{2})=(\frac{a+1}{2},\frac{0+2}{2})$

$\Rightarrow (1,\frac{b-1}{2})=(\frac{a+1}{2},1)$

$\Rightarrow \frac{a+1}{2}=1and\frac{b-1}{2}=1$

$\Rightarrow a+1=2andb-1=2$

$\Rightarrow a=1andb=3$

Formula: 1 Mark
Application: 1 Mark
Answer: 1 Mark

Let the given points be A(4, – 1), B(6, 0), C(7, 2) and D(5, 1) respectively. Then,

Coordinates of the mid-point of AC are

$(\frac{4+7}{2},\frac{-1+2}{2})=(\frac{11}{2},\frac{1}{2})$ 

Coordinates of the mid-point of BD are

$(\frac{6+5}{2},\frac{0+1}{2})=(\frac{11}{2},\frac{1}{2})$ 

Thus, AC and BD have the same mid-point.

Hence, ABCD is a parallelogram.

Now
Distance between the points is given by
$\sqrt{{(x_1-x_2)}^2+{(y_1-y_2)}^2}$ 
So, 
AB = $\sqrt{(6-4{)}^2+(0+1{)}^2}=\sqrt{5}$ 

BC = $\sqrt{(7-6{)}^2+(2-0{)}^2}=\sqrt{5}$ 

$\therefore$ AB = BC 

So, ABCD is a parallelogram whose adjacent sides are equal.

$\Rightarrow$  ABCD is a rhombus.

We have, 

AC = $\sqrt{(7-4{)}^2+(2+1{)}^2}=3\sqrt{2}$   and , 

BD = $\sqrt{(6-5{)}^2+(0-1{)}^2}=\sqrt{2}$ 

 
Clearly, the diagonals AC ≠ BD.

So, ABCD is not a square. 
 

Formula: 1 Mark
Concept: 1 Mark
Answer: 2 Marks

Let D, E, F be the mid-points of the sides BC, CA and AB respectively.

Then, the coordinates of D, E and F are

$D(\frac{5+3}{2},\frac{3-1}{2})=D(4,1),$

$E(\frac{3+7}{2},\frac{-1-3}{2})=E(5,-2)$

and $F(\frac{7+5}{2},\frac{-3+3}{2})=F(6,0)$

Distance between the points is given by
$\sqrt{{(x_1-x_2)}^2+{(y_1-y_2)}^2}$
$\therefore AD=\sqrt{(7-4{)}^2+(-3-1{)}^2}$

$\Rightarrow AD=\sqrt{9+16}=5units$

$BE=\sqrt{(5-5{)}^2+(-2-3{)}^2}$

$\Rightarrow BE=\sqrt{0+25}=5units$

and, $CF=\sqrt{(6-3{)}^2+(0+1{)}^2}$

$\Rightarrow CF=\sqrt{9+1}=\sqrt{10}units$

Concept : 1 Mark
Application : 1 Mark
Calculation : 2 Marks

Consider a  parallelogram ABCD .

Let O be the point of intersection of the diagonals AC and BD

We know that diagonals of a parallelogram bisect each other.

$\Rightarrow$ O is the midpoint of both the digonals AC and BD

Now, coordinates of O as mid-point of BD are

$O(x,y)=\left(\frac{x_1+x_2}{2}\right),\left(\frac{y_1+y_2}{2}\right)$

$\Rightarrow O(a,b)=\frac{4+3}{2},\frac{y+5}{2}\dots \dots \left(1\right)$ 

Also, coordinates of O as mid-point of AC are

$O(a,b)=\frac{1+x}{2},\frac{2+6}{2}\dots \dots \left(2\right)$

From (1) and (2), we have

$\frac{4+3}{2}=\frac{1+x}{2}$

$\frac{7}{2}=\frac{1+x}{2}\Rightarrow x=6$

And $\frac{y+5}{2}=\frac{2+6}{2}$

$\frac{y+5}{2}=\frac{8}{2}\Rightarrow y=3$

Formula for centroid and mid-point: 1 Mark each

Let A$(x_1,y_1)$,B$(x_2,y_2)$ and C$(x_3,y_3)$ be the vertices of triangle ABC.

Let P(1,1), Q(2,-3), R(3,4) be the mid-points of sides AB, BC and CA respectively.

Then, P is the mid-point of BC

$\Rightarrow \frac{x_1+x_2}{2}=1,\frac{y_1+y_2}{2}=1$

$\Rightarrow x_1+x_2=2$ and $y_1+y_2=2\dots \dots \left(1\right)$

Q is the mid-point of AB

$\Rightarrow \frac{x_2+x_3}{2}=2,\frac{y_2+y_3}{2}=-3$

$\Rightarrow x_2+x_3=4$ and $y_2+y_3=-6\dots \dots \left(2\right)$

R is the mid-point of AC

$\Rightarrow \frac{x_1+x_3}{2}=3,\frac{y_1+y_3}{2}=4$

$\Rightarrow x_1+x_3=6$ and $y_1+y_3=8\dots \dots \left(3\right)$

Adding (1),(2) and (3) we get,

$x_1+x_2+x_2+x_3+x_1+x_3=2+4+6$

and $y_1+y_2+y_2+y_3+y_1+y_3=2-6+8$

$\Rightarrow x_1+x_2+x_3=6\dots \dots \left(4\right)$ 

and $y_1+y_2+y_3=2\dots \dots \left(5\right)$

The coordinates of the centroid of $△ABC$ are

$(\frac{x_1+x_2+x_3}{3},\frac{y_1+y_2+y_3}{3})=\frac{6}{3},\frac{2}{3}=2,\frac{2}{3}$ [Using (4) and (5)]

Formula: 1 Mark
Calculations: 2 Marks
Answer: 1 Mark

For points A,B,C to be collinear, area of triangle ABC formed by these points should be zero.

$\Rightarrow$ The area of  $\Delta ABC=0$

 $\Rightarrow \frac{1}{2}\left[x_1\right(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2\left)\right]=0$

$\Rightarrow \frac{1}{2}\left[k\right(2k-6+2k)+(–k+1\left)\right(6–2k-2+2k)+(–4–k\left)\right(2–2k-2k\left)\right]=0$

$\Rightarrow \frac{1}{2}\left[k\right(4k-6)+(–k+1\left)\right(4)+(–4–k\left)\right(2–4k\left)\right]=0$

$\Rightarrow \frac{1}{2}[4k^2-6k-4k+4-8+16k-2k+4k^2]=0$

$\Rightarrow 8k^2+4k-4=0$

$\Rightarrow 8k^2+4k-4=0$

$\Rightarrow 2k^2+k-1=0$

$\Rightarrow (k+1)(2k-1)=0$

$\Rightarrow k=-1ork=\frac{1}{2}$

Concept : 1 Mark
Application : 1 Mark
Calculation : 2 Marks

Let ABCD be a square and two opposite vertices of it are A(-1, 2) and C(3, 2). ABCD is a square.

Since ABCD is a square.

$\Rightarrow AB=BC$

$\Rightarrow A{B}^2=B{C}^2$
[Distance between the points is given by
$\sqrt{{(x_1-x_2)}^2+{(y_1-y_2)}^2}$]

$\Rightarrow {(x+1)}^2+{(y-2)}^2={(x-3)}^2+{(y-2)}^2$

$\Rightarrow x^2+2x+1=x^2-6x+9$

$\Rightarrow 2x+6x=9-1=8$

$\Rightarrow 8x=8\Rightarrow x=1$

ABC is right $\Delta$ at B, then

$A{C}^2=A{B}^2+B{C}^2$ (Pythagoras theorem)

$\Rightarrow (3+1{)}^2+(2-2{)}^2=(x+1{)}^2+(y-2{)}^2+(x-3{)}^2+(y-2{)}^2$

$\Rightarrow 16=2(y-2{)}^2+(1+1{)}^2+(1-3{)}^2$

$\Rightarrow 16=2(y-2{)}^2+4+4\Rightarrow 2(y-2{)}^2=16-8=8$

$\Rightarrow (y-2{)}^2=4\Rightarrow y-2=\pm 2\Rightarrow y=4$ and 0

i.e when x = 1 then y = 4 and 0

Co-ordinates of the opposite vertices are: B(1,0) or D(1,4)

Concept : 1 Mark
Application : 1 Mark
Calculation : 2 Marks

Area of quadrilateral $ABCD=|Areaof\Delta ABC|+|Areaof\Delta ACD|$

We have,

$\therefore$ Area of $\Delta ABC=\frac{1}{2}|(1\times -3+7\times 2+12\times 1)-(7\times 1+12\times (-3)+1\times 2)|$

$\Rightarrow \Delta ABC=\frac{1}{2}|(-3+14+12)-(7-36+2)|$

$\Rightarrow \Delta ABC=\frac{1}{2}|23+27|=25sq.units$

Also, we have

$\therefore$ Area of $\Delta ACD=\frac{1}{2}|(1\times 2+12\times 21+7\times 1)-(12\times 1+7\times 2+1\times 21)|$

 $\Rightarrow \Delta ACD=\frac{1}{2}|(2+252+7)-(12+14+21)|$

$\Rightarrow \Delta ACD=\frac{1}{2}|261-47|=107sq.units$

$\therefore$ Area of quadrilateral ABCD = 25 + 107 = 132 sq. units