Coordinate Geometry Test

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Coordinate Geometry Test - 28

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Weekly Quiz Competition

Question 1
1 / -0

Find the area of the triangle whose vertices are $$(a, b + c), (a, b - c)$$ and $$(-a, c)$$.

A
$$2ac$$

B
$$2bc$$

C
$$b(a + c)$$

D
$$c (a - b)$$

Solution

Area of a triangle $$ (A) =\left| \cfrac { {x }_{ 1 }({ y }_{ 2 }-{ y }_{ 3 })+{ x }_{ 2 }({ y }_{ 3 }-{ y }_{ 1 })+{ x }_{3 }({ y }_{ 1 }-{ y }_{ 2 }) }{ 2 } \right| $$

Hence, substituting the points $$({ x }_{ 1 },{ y }_{ 1 }) = (a,b+c) $$ , $$({ x }_{ 2

},{ y }_{ 2 }) = (a,b-c) $$ and $$({ x }_{ 3 },{ y }_{ 3 }) = (-a,c) $$

$$ A=\left| \cfrac { a(b-c-c)+a(c-b-c)-a(b+c-b+c) }{ 2 } \right| $$
$$=\left| \cfrac { a(b-2c)+a(-b)-a(2c) }{ 2 } \right|$$
$$ =\left| \cfrac { ab-2ac-ab-2ac }{ 2 } \right| $$
$$=\left| \cfrac { -4ac }{ 2 } \right| $$
$$=2ac$$ square units
Question 2
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The area of the triangle whose vertices are $$A(1,1), B(7, 3)$$ and $$C(12, 2)$$ is

A
$$25$$ square units

B
$$8$$ square units

C
$$16$$ square units

D
$$12$$ square units

Solution

We know that the area of the triangle whose vertices are $$\displaystyle (x_{1},y_{1}),(x_{2},y_{2}),$$ and $$(x_{3},y_{3})$$ is $$\cfrac{1}{2}\left | x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1}) +x_{3}(y_{1}-y_{2})\right |$$
$$\therefore$$
Area $$= \dfrac{1}{2} [1(1)+ 7(1)+ 12(-2)] = -8$$, but area can not be negative
Hence, area = 8 square units.
Question 3
1 / -0

The mid-point of the line segment joining $$( 2a, 4)$$ and $$(-2, 2b)$$ is $$(1, 2a + 1 )$$. The values of $$a$$ and $$b$$ are

A
$$a = b, b = -1$$

B
$$a = 2, b = -3$$

C
$$a = 3, b = - 2$$

D
$$a =2, b = 3$$

Solution

Midpoint of two points $$ =\left( \cfrac { { x }_{ 1 }+{ x}_{ 2 } }{ 2 } ,\cfrac { { y }_{ 1 }+y_{ 2 } }{ 2 } \right) $$
Given, midpoint of $$ (2a,4) $$ and $$ (-2,2b) = (1,2a+1) $$
$$ => \left(\cfrac { 2a-2 }{ 2 } ,\cfrac { 4+2b }{ 2 }\right)= (1,2a+1) $$
$$ => \cfrac { 2a-2 }{ 2 } = 1 ; \cfrac { 4+2b }{ 2 } = 2a + 1 $$
$$ => 2a -2 = 2 $$
$$=> a = 2 $$

And, $$ \cfrac { 4+2b }{ 2 } = 2a + 1 $$
$$=> \cfrac { 4+2b }{ 2 } = 2(2) + 1 = 5 $$
$$ => 4 + 2b = 10 $$
$$ => 2b = 6 $$
$$=> b = 3 $$
Question 4
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Find the co ordinates of the point which divides the line segment joining the points (6, 3) and (-4, 5) in the ratio 3 : 2 internally

A
$$\displaystyle \left ( 0,\frac{21}{5} \right )$$

B
$$\displaystyle \left ( 1,\frac{21}{5} \right )$$

C
$$\displaystyle \left (0, \frac{22}{5} \right )$$

D
$$\displaystyle \left ( 2,\frac{21}{5} \right )$$

Solution

Let $$P(x, y)$$ be the point which divides the line segment $$AB$$ in the ratio 3:2 internally

$$ We\quad know\quad that\quad if\quad a\quad point\quad P\left( x,y \right) \quad divides\quad the\quad line\quad segment\quad AB\quad \\ joining\quad the\quad points\quad A\left( { x }_{ 1 },{ y }_{ 1 } \right) \quad \& \quad B\left( { x }_{ 2 },{ y }_{ 2 } \right) \quad in\quad the\quad ratio\quad m:n\\ then,\quad by\quad the\quad section\quad formula,\quad $$

$$\Rightarrow$$ $$P=\left(\dfrac{mx_2+nx_1}{m+n},\,\dfrac{my_2+n_1}{m+n}\right)$$

$$=\left(\dfrac{(3)(-4)+(2)(6)}{3+2},\,\dfrac{(3)(5)+(2)(3)}{3+2}\right)$$

$$=\left(0,\dfrac{21}{5}\right)$$
.
Question 5
1 / -0

$$(3, 1), (-3, 2)$$ and $$\displaystyle (0,2-\sqrt{3})$$ are the vertices of __________ triangle of area ___________.

A
an isosceles, $$81$$ sq. units

B
a scalene, $$\cfrac {-3+6\sqrt{3}}{2}$$ sq. units

C
an equilateral, $$\displaystyle 9\sqrt{3}$$ sq. units

D
a right angled, $$81$$ sq. units

Solution

Distance between two points $$= \sqrt { \left( { x }_{ 2 }-{x }_{ 1 } \right) ^{ 2 }+\left( { y }_{ 2 }-{ y }_{ 1 } \right) ^{ 2 } } $$
Distance between the points $$A (3,1) $$ and $$B (-3,2) = \sqrt { \left( -3-3 \right) ^{ 2 }+\left( 2 - 1 \right) ^{ 2 } } = \sqrt { 36 + 1 } = \sqrt { 37 }$$
Distance between the points $$B(-3,2) $$ and $$C (0,2-\sqrt {3}) = \sqrt { \left( 0 + 3 \right) ^{ 2 }+\left(2 - \sqrt {3} - 2\right) ^{ 2 } } = \sqrt { 9 + 3 } = \sqrt { 12 } $$
Distance between the points $$A(3,1) $$ and $$ C(0,2-\sqrt {3})$$
$$ = \sqrt { \left( 0-3\right) ^{ 2 }+\left( 2-\sqrt {3} - 1\right) ^{ 2 } } = \sqrt { 9 + 1 + 3 -2\sqrt {3} } = \sqrt { 13 - 2\sqrt {3} } $$
Since the length of the sides between all vertices are different, they are the vertices of a scalene triangle.
Area of a triangle $$= \left| \cfrac { {x }_{ 1 }({ y }_{ 2 }-{ y }_{ 3 })+{ x }_{ 2 }({ y }_{ 3 }-{ y }_{ 1 })+{ x }_{3 }({ y }_{ 1 }-{ y }_{ 2 }) }{ 2 } \right| $$

Area of $$\triangle ABC= \left| \cfrac { (3)(2-2+\sqrt {3})+(-3)(2-\sqrt {3} - 1)+0(1-2) }{ 2 } \right| $$
$$ = \left| \cfrac { 3\sqrt {3} -3 + 3\sqrt {3} }{ 2 } \right| $$
$$ = \cfrac{-3 + 6 \sqrt {3}}{2}$$ sq. units
Question 6
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Three points A, B and C have coordinates $$(a, b + c), \ (b, c + a)$$ and $$(c, a + b)$$, respectively. The area of the triangle ABC will be:

A
$$\displaystyle a^{2}+b^{2}+c^{2}$$

B
$$\displaystyle \dfrac{a^{2}+b^{2}+c^{2}}2$$

C
$$\displaystyle \dfrac{a^{2}+b^{2}+c^{2}}4$$

D
$$0$$

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 the points: $$A(a, b+c) , B(b, c+a) , C(c, a+b)$$
Therefore, area:

$$A =\dfrac{ 1 }{ 2 }\left[ a(c+a-a-b)+b(a+b-b-c)+c(b+c-c-a) \right]$$

$$A =\dfrac{ 1 }{ 2}{\left[ a(c-b)+b(a-c)+c(b-a) \right] }$$

$$A=\dfrac{ 1 }{ 2}{\left[ ac-ab+ba-bc+cb-ac \right] }=0$$
Question 7
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If the coordinates of two points A and B are $$(3, 4)$$ and $$(5, -2)$$, respectively, then the coordinates of any point P if $$PA = PB$$ and area of $$\displaystyle \Delta PAB=10$$ is

A
$$(7, 2)$$ or $$(1, 0)$$

B
$$(-7, 2)$$ or $$(3, 0)$$

C
$$(7, -2)$$ or $$(5, 0)$$

D
$$(7, -2)$$ or $$(-1, 0)$$

Solution

Let P $$ = (x, y) $$
Given, $$ PA = PB $$
$$=> {PA}^{2} = {PB}^{2} $$
$$ => {(x-3)}^{2} + {(y-4)}^{2} = {(x-5)}^{2} + {(y+2)}^{2} $$
$$=> {x} ^{2} -6x + 9 + {y} ^{2} -8y + 16 = {x}^{2} -10x + 25 + {y}^{2} + 4y + 4 $$
$$=> 4x - 12y = 4 $$
$$=> x - 3y = 1 $$ .......(i)

Also, Area of $$ \Delta PAB=10 $$
$$\left| \cfrac {x (4+2)+3(-2-y)+5(y-4)}{ 2 } \right|=10 $$
$$ \left| \cfrac { 6x -6-2y +5y -20 }{ 2 } \right| = 10 $$
$$ \cfrac {6x+3y -26}{2} = \pm 10 $$
$$ 6x+2y-26= \pm 20 $$
$$ 6x+2y = 46 $$ ........(ii)
or $$ 6x + 2y = 6 $$ .......(iii)
Solving equation (i) and (ii), we get $$ x = 7, y = 2 $$
Solving equation (i) and (iii), we get $$ x = 1, y = 0 $$.
So, the co-ordinates of P are $$ (7, 2)$$ or $$(1, 0) $$.
Question 8
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Find the co-ordinates of the point dividing the joining of line segment $$A(1, -2)$$ and $$B(4, 7)$$ which divides $$AB$$ internally in the ratio $$1 : 2.$$

A
$$(2, 1)$$

B
$$(-2,-3)$$

C
$$(1,2)$$

D
$$(2,3)$$

Solution

The given points are $$A(1,-2)$$ and $$B(4,7)$$.
We have to find the coordinate of points which divide the line segment internally in the ratio $$1:2$$
Let the point $$C$$ divides the line segment $$AB$$ in the ratio $$1:2$$

$$ We\quad know\quad that\quad if\quad a\quad point\quad C\left( x,y \right) \quad divides\quad the\quad line\quad segment\quad AB\quad \\ joining\quad the\quad points\quad A\left( { x }_{ 1 },{ y }_{ 1 } \right) \quad \& \quad B\left( { x }_{ 2 },{ y }_{ 2 } \right) \quad in\quad the\quad ratio\quad m:n\\ then,\quad by\quad the\quad section\quad formula,\quad $$

$$\Rightarrow$$ $$C=\left(\dfrac{mx_2+nx_1}{m+n},\,\dfrac{my_2+n_1}{m+n}\right)$$

$$=\left(\dfrac{(1)(4)+(2)(1)}{1+2},\,\dfrac{(1)(7)+(2)(-2)}{1+2}\right)$$

$$=\left(\dfrac{6}{3},\dfrac{3}{3}\right)$$

$$=(2,1)$$
Question 9
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The triangle with vertices A(4, 4), B(-2, -6) and C(4, -1) is shown in the diagram. The area of $$\Delta$$ ABC is _______

A
5 sq. units

B
12 sq. units

C
15 sq. units

D
20 sq. units

Solution

We know that the area of the triangle whose vertices are $$\displaystyle (x_{1},y_{1}),(x_{2},y_{2}),$$ and $$(x_{3},y_{3})$$ is $$= \cfrac{1}{2}\left | x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1}) +x_{3}(y_{1}-y_{2})\right |$$

$$=\dfrac{1}{2}|4(-6-(-1))-2(-1-4)+4(4-(-6))|$$

$$=\dfrac{1}{2}|4(-5)-2(-5)+4(10)|$$

$$=\dfrac{1}{2}|30|$$

$$=15 $$ sq. units
Question 10
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The coordinates of the points P and Q are respectively (4, - 3) and (-1, 7) Then the abscissa of a point R on the line segment PQ such that $$\displaystyle \frac{PR}{PQ}=\frac{3}{5}$$ is

A
$$\displaystyle \frac{18}{5}$$

B
$$\displaystyle \frac{17}{5}$$

C
$$\displaystyle \frac{17}{8}$$

D
1

Solution

Using the section formula, if a point $$(x,y)$$ divides the line joining the

points $$({ x }_{ 1 },{ y }_{ 1 })$$ and $$({ x }_{ 2 },{ y }_{ 2 })$$ internally

in the ratio $$ m:n $$, then $$(x,y) = \left( \frac { m{ x }_{ 2 } + n{ x }_{ 1

} }{ m + n } ,\frac { m{ y }_{ 2 } + n{ y }_{ 1 } }{ m + n }

\right) $$

Substituting $$({ x }_{ 1 },{ y }_{ 1 }) = (4,-3) $$ and $$({x }_{ 2 },{ y }_{

2 }) = (-1,7) $$ and $$ m = 3, n = 2 $$ in the section formula, we get

the abscissa of the point $$ ( \frac { 3(-1) + 2(4) }{ 3+2 }) =1 $$

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Coordinate Geometry Test - 28

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Coordinate Geometry Test - 28

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Question 1

$$2ac$$

$$2bc$$

$$b(a + c)$$

$$c (a - b)$$

Solution

Question 2

$$25$$ square units

$$8$$ square units

$$16$$ square units

$$12$$ square units

Solution

Question 3

$$a = b, b = -1$$

$$a = 2, b = -3$$

$$a = 3, b = - 2$$

$$a =2, b = 3$$

Solution

Question 4

$$\displaystyle \left ( 0,\frac{21}{5} \right )$$

$$\displaystyle \left ( 1,\frac{21}{5} \right )$$

$$\displaystyle \left (0, \frac{22}{5} \right )$$

$$\displaystyle \left ( 2,\frac{21}{5} \right )$$

Solution

Question 5

an isosceles, $$81$$ sq. units

a scalene, $$\cfrac {-3+6\sqrt{3}}{2}$$ sq. units

an equilateral, $$\displaystyle 9\sqrt{3}$$ sq. units

a right angled, $$81$$ sq. units

Solution

Question 6

$$\displaystyle a^{2}+b^{2}+c^{2}$$

$$\displaystyle \dfrac{a^{2}+b^{2}+c^{2}}2$$

$$\displaystyle \dfrac{a^{2}+b^{2}+c^{2}}4$$

$$0$$

Solution

Question 7

$$(7, 2)$$ or $$(1, 0)$$

$$(-7, 2)$$ or $$(3, 0)$$

$$(7, -2)$$ or $$(5, 0)$$

$$(7, -2)$$ or $$(-1, 0)$$

Solution

Question 8

$$(2, 1)$$

$$(-2,-3)$$

$$(1,2)$$

$$(2,3)$$

Solution

Question 9

5 sq. units

12 sq. units

15 sq. units

20 sq. units

Solution

Question 10

$$\displaystyle \frac{18}{5}$$

$$\displaystyle \frac{17}{5}$$

$$\displaystyle \frac{17}{8}$$

1

Solution

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