Coordinate Geometry Class 10 MCQ with Solution Maths

Coordinate Geometry Class 10 MCQ with Solution

1.  If  two vertices of an equilateral  triangle be (0,0), (3,\,\sqrt 3 ), find the third vertex.

(A) (2\sqrt 3 ,\,0)                                  

(B)  (0,\,\,2\sqrt 3 )         

(C) (3,\, - \sqrt 3 )           

(D) Both (B) and (C)

Ans.     (D)

Sol.      ΔACD and ΔABC are equilateral triangles.

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2.   Find the value of a if a line 3x + 4y = a passes through the point of intersection of line 3x – 5y = 1 and line
5x + 2y = 12 :

(A) – 6 

(B) 8    

(C) 10  

(D) 12

Ans.     (C)

Sol.      3x – 5y = 1                       …(1)

            5x + 2y = 12                     …(2)

            point of  intersection (1) and (2) is (2, 1)

            Now  3(2) + 4(1) = 10

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3.   If  3x + ay = 4 and bx – 3y = 13 passes through (2, –1) then find a + b :

(A) 12 

(B) 11 

(C) 7    

(D) 13

Ans.     (C)

Sol.      3(2) + a(–1) = 4  ⇒ a = 2

            2b + 3 = 13  ⇒ b = 5

            Now  a + b = 2 + 5 = 7

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4.    If ax + by + 21 = 0 and –2x –by = 5a + 2 passes through (–3,2) then find the value of a and b :

(A) 5, 2

(B) –5,3                                       

(C)  \frac{{25}}{4}\,,\,\frac{{92}}{{16}}       

(D) \frac{{25}}{8}\,,\,\frac{{ - 93}}{{16}}

Ans.     (D)

Sol.      a(–3) + b(2) + 21 = 0         …(1)

            –2(–3) – b(2) = 5a + 2        …(2)

            Solving (1) and (2)

a = \frac{{25}}{8},\,\,b =  - \frac{{93}}{{16}}

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5.   Which of the following line is not parallel to 3x – 5y = 7:

(A) 6x – 10y = 3                           

(B) – 6x + 10y –7 = 0                   

(C) 5x - \frac{{25}}{3}y\, = \,1  

(D) 4x – 7y = 8

Ans.     (D)

Sol.      4x – 7y = 8 is not parallel to 3x – 5y = 7 because for these equations\frac{{{a_1}}}{{{a_2}}} \ne \,\frac{{{b_1}}}{{{b_2}}} which is the condition for unique solution.

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6.   If a line 3x – 5y = b passes through the intersection point of 5x – 4y = –2 and 3x + 4y = 18 then find value of b :

(A) 6    

(B) 7    

(C) 9    

(D) –9

Ans.     (D)

Sol.      point of intersection 5x – 4y = – 2 and 3x + 4y = 18 is (2, 3).

            Line 3x – 5y = b passes through (2, 3) so 3(2) – 5(3) = b = – 9

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7.    Find the number of points of  intersection lines –3x + 4y = 0, 4x – 5y = 1 and 2x + 3y = 17 :

(A) 0    

(B) 3    

(C) 2    

(D) 1

Ans.     (D)

Sol.      Intersection of – 3x + 4y = 0 and 4x – 5y = 1 is (4, 3) and 2x + 3y = 17 passes through this point. So point of intersection is one only.

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8.   Find  the length of  the line segment made on line 9x + 12 y = 108 by  both the axes ::

(A) 9 units                                   

(B) 15 units                                 

(C) 12 units                                 

(D) 98 units

Ans.     (B)

Sol.      9x + 12y = 108               

\frac{x}{{12}} + \frac{y}{9} = 1

            AB =\sqrt {{9^2} + {{12}^2}}  = 15

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9.   In what ratio does the x-axis divide the line segment joining the points (–3, 0) and (3, 0) :

(A) 3 : 1                                      

(B) 1 : 2                                      

(C) 2 : 1                                      

(D) 1 : 1

Ans.     (D)

Sol.      1 : 1 because origin is the mid point of the line segement.

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10.   The vertices of a triangle  are  (3,–1), (–3, –1) and (6, 2) then coordinates of its centroid is :

(A) (0, 2)                                     

(B) (2, 0)                                     

(C) (2, 4)                                     

(D) (4, 2)

Ans.     (B)

Sol.      x\, = \,\frac{{{x_1}\, + \,{x_2}\, + \,{x_3}}}{3} and y\, = \,\frac{{{y_1}\, + \,{y_2}\, + \,{y_3}}}{3}

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11.   If points (a, 0), (0, b) and (1, 1) are collinear, then \frac{1}{a} + \frac{1}{b}=

(A) 1    

(B) 2

(C) 0    

(D) –1

Ans.     (A)

Sol.      A(a, 0), B(0, b) & C(1, 1) are collinear

⇒      \frac{{b\, - \,0}}{{0\, - \,a}}\,\, = \,\,\frac{{1\, - \,b}}{{1\, - \,0}}\,\, \Rightarrow \,\,\frac{b}{{ - a}}\, = \,\frac{{1\, - \,b}}{1}\,\,

⇒     \frac{1}{{ - a}}\,\, = \,\frac{{1\, - \,b}}{b}

⇒       \frac{1}{a}\, + \,\frac{1}{b}\,\, = \,\,1

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12.   The points (0, 8), (1, 8) and (1, – 9/8) are the vertices of –

(A) An equilateral triangle                                 

(B) The collinear points

(C) An isosceles triangle                                  

(D) A right – angled triangle

Ans.     (D)

Sol.      AB = \sqrt 1  = 1,\,\,BC = \left( {\frac{{73}}{8}} \right),\,\,\,AC = \sqrt {1 + \frac{{{{73}^2}}}{{64}}}  = \frac{{\sqrt {64 + {{73}^2}} }}{8}

⇒     AC² = AB² + BC²  ⇒ ΔABC is a right angle triangle

⇒    Option ‘D’ correct.

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13.   If the points (–2, –1), (1, 0), (x, 3) and (1, y) form a parallelogram, find the value of x² – y :

(A) 12  

(B) 14

(C) 13  

(D) 15

Ans.     (B)

Sol.      A(–2, –1), B(1, 0), C(2, 3) & D(1, y) form a parallelogram

⇒     mid point of diagonal AC =\left( {\frac{{x - 2}}{2},1} \right)

&       mid point of diagonal BD =\left( {1,\,\,\frac{y}{2}} \right)

⇒     \frac{{x - 2}}{2} = 1\,\,\& \,\,\frac{y}{2} = 1\,\,\, \Rightarrow \,\,x = 4,\,\,y = 2

⇒     x² – y = 16 – 2 = 14    ⇒ Option ‘B’ correct

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14.   The points (a, a) (– a, – a) and (\ - \sqrt 3 a, \sqrt 3 a) form the vertices of an :

(A) Scalene triangle                                         

(B) Right angled triangle                                  

(C) Isosceles right angled triangle                    

(D) Equilateral triangle                                      

Ans.     (D)

Sol.      The points A(a, a), B(–a, –a), C( - \sqrt 3 a,\,\,\sqrt 3 a)

⇒     AB = \sqrt {4{a^2} + 4{a^2}}  = 2a\sqrt 2 ,BC = 2a\sqrt 2   , CA = 2a\sqrt 2

⇒       ΔABC is an equilateral triangle.

⇒     Option ‘D’ correct.

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15.   The centroid of the triangle whose vertices are (4, –8), (–9, 7) and (8, 13) is :

(A) (1, 4)                                                         

(B) (1, 3)

(C) (1, 5)                                                         

(D) (1, 9)                                                         

Ans.     (A)

Sol.      centroid =\left( {\frac{{4 - 9 + 8}}{3},\frac{{ - 8 + 7 + 13}}{3}} \right)

                                      = (1, 4)

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16.   The ratio in which the line segment joining (3, 4) and (–1, 2) is divided by the y-axis is :

(A) 1 : 2                                                          

(B) 1 : 3

(C) 3 : 2                                                          

(D) None of these                                           

Ans.     (B)

Sol.      Let y axis divides them in m : n, x coordinate at y-axis will be zero.

            so, \frac{{3n + m( - 1)}}{{m + n}} = 0

            so, \frac{m}{n} =\frac{1}{3}

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17.  The point (–5, –2) lies in which quarant :

(A)  Ist 

(B)  IInd

(C)  IIIrd

(D) IVth                                                           

Ans.     (C)                                                      

Sol.      (–5, –2) lies in IIIrd quadrant.

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18.   The point (a, a) lies in which quadrant where (a > 0) :

(A)  Ist 

(B)  IInd

(C)  IIIrd

(D)  IVth

Ans.     (A)

Sol.      (a, a) lies in Ist quadrant.

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19.   The point (a, –a) lies in which quadrant where (a < 0) :

(A)  Ist 

(B)  IInd

(C)  IIIrd

(D) IVth

Ans.     (B)

Sol.      (a, –a) lies in IInd quadrant.

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20.   The point (–a, –a) lies in which quadrant where (a < 0) :

(A)  Ist 

(B)  IInd

(C)  IIIrd

(D)  IVth

Ans.     (A)

Sol.      (–a, –a) lies in Ist quadrant.

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21.   The slope of any line parallel to X-axis is :

(A) 0                                              

(B) 1

(C) – 1                                           

(D) Not defined                            

Ans.     (A)

Sol.      Slope of x-axis is ‘zero(0)’.

‡ Slopes of two parallel lines are same

⇒ Slope of a line parallel to x-axis is also ‘zero’.

⇒         Option ‘A’ correct.

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22.  The distance between the points (cosθ, sinθ) and (sinθ, –cosθ) is :

(A)\sqrt 3                                                                       

(B) \sqrt 2

(C) 2                                              

(D) 1

Ans.     (B)

Sol.      Let P≡ (cos θ, sin θ), Q ≡ (sin θ, –cos θ)

            PQ =\sqrt {{{(\cos \theta  - \sin \theta )}^2} + {{(\sin \theta  + \cos \theta )}^2}} PQ = \sqrt {2 + 0}  = \sqrt 2

⇒    Option ‘B’ is correct.

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23.   (9, 9) lies in which quadrant :

(A)  Ist                                           

(B)  IInd

(C)  IIIrd                                         

(D)  IVth                                       

Ans.     (A)

Sol.      (9, 9) lies in Ist quadrant.

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24.   (–3, –3) lies in which quadrant :

(A)  Ist                                           

(B)  IInd

(C)  IIIrd                                         

(D)  IVth                                        

Ans.     (C)

Sol.      (–3, –3) lies in IIIrd quadrant.

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25.   If points (a, 0), (0, b) and (2, 2) are collinear, then\frac{1}{a} + \frac{1}{b} = :

(A)  1                                           

(B)  2

(C)  0                                             

(D) \frac{1}{2}

Ans.     (D)

Sol.      Equation of line through (a, 0), (0, b) and (2, 2) lies on it.

     \frac{x}{a} + \frac{y}{b} = 1

so, \frac{2}{a} + \frac{2}{b} = 1

     \frac{1}{a} + \frac{1}{b} = =\frac{1}{2}

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26.   Find the controid of triangle whose vertices are (0, 0), (0, 1) & (2, 0) :

(A) \left( {\frac{2}{3},\frac{1}{3}} \right)                                                          

(B)  (0, 0)

(C)  (1, 5)                                                        

(D) None of these

Ans.     (A)

Sol.      Controid  =\left( {\frac{{0 + 0 + 2}}{3},\frac{{0 + 1 + 0}}{3}} \right) =\left( {\frac{2}{3},\frac{1}{3}} \right)

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27.   If  the distance between A (x, y) and B( 3,2) is \sqrt 2 find the value of x and y. If 2x + y = 7 :

(A)  (2, 3)                                                        

(B) \left( {\frac{{16}}{5},\frac{3}{5}} \right)

(C) Both (A) and (B)                                        

(D) None of these

Ans.     (C)

Sol.      ‡ d(AB) = \sqrt 2

⇒    \sqrt {{{(x - 3)}^2} + {{(y - 2)}^2}}  = \sqrt 2  (using distance formula)

⇒     (x – 3)² + (y – 2)² = 2

⇒      x² + 9 – 6x – 4y + 11 = 0

⇒     x² + y² – 6x – 4y + 11 = 0

also    2x + y = 7  ⇒ y = 7 – 2x

∴         x² + (7 – 2x)² – 6x – 4(7 – 2x) + 11 = 0

⇒         5x² – 26x + 32 = 0

⇒      x = \frac{{26 \pm 6}}{{10}} = \frac{{32}}{{10}},2                        

or    x = \frac{{16}}{5},2\,\,\,\, \Rightarrow y = 3,\,\,\frac{3}{5}                            

∴     The two possible ordered pairs of (x, y) are \left( {\frac{{16}}{5},\,\,\frac{3}{5}} \right)\,\,\& \,\,\left( {2,\,\,3} \right)

            Hence correct option ‘C’

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28.   The distance of the point (x, y) from Y-axis is :

(A) x                                              

(B) y    

(C) | x |                                          

(D) | y |

Ans.     (C)

Sol.      Distance of point P(x, y) from y-axis is | x | units. 

            Option ‘C’ correct.

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29.   Find the area bounded by lines x + y = 2 and x – y = 2 and y-axis

(A) 4 square units                           

(B)  square units

(C) (3 ¸ 3 ¸ 3 + 3 – 3 × 2 + 7 –\frac{1}{3} ) square units

(D) All of these

Ans.     (A)

Sol.      Area bounded by lines x + y = 2 &

                                             x – y  = 2 and y-axis will equal to 4.

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30.   The distance between the points (a cos θ + b sin θ, 0) and (0, a sin θ – b cos θ) is

(A) a² + b²                                                       

(B) a + b

(C) a² – b²                                                       

(D)\sqrt {{a^2} + {b^2}}

Ans.     (D)

Sol.      distance  = \sqrt {{{(a\cos \theta  + b\sin \theta  - 0)}^2} + {{(a\sin \theta  - b\cos \theta )}^2}}

                             = \sqrt {{a^2} + {b^2}}

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31.  (7, –7) lies in which quadrant :

(A) 1st                                           

(B) 2nd

(C) 3rd                                           

(D) 4th

Ans.     (D)

Sol.      (7, –7) lies in in the 4th quadrant

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32.   The point of intersection of X and Y axis is called :

(A) origin                                                        

(B) null point                                 

(C) Common point                                           

(D) None of these

Ans.     (A)

Sol.      The point of intersection of x-axis and y-axis is called origin.

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33.  The point (2, 3) is at a distance of _______ units from x-axis :

(A) 2                                              

(B) 5

(C) 3                                              

(D) None of these                         

Ans.     (C)

Sol.      The point (2, 3) is at a distance of 3 units from x-axis.

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34.   The point (3, 2) is at a distance of ____________ units from y-axis :

(A) 2                                              

(B) 3

(C) 5                                             

(D) None of these                         

Ans.     (B)

Sol.      The point (3, 2) is at a distance of 3 units from y-axis.

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35.   The point (– 3, 2) belongs to :

(A) Ist quadrant                                               

(B) IInd quadrant                           

(C) IIIrd quadrant                                             

(D) IVth quadrant                          

Ans.     (B)

Sol.      The point (–3, 2) belongs to IInd quadrant.

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36.   The point (0, – 2) lies on :

(A) + ve X-axis                                                

(B) + ve Y-axis

(C) – ve X-axis                                                

(D) – ve Y-axis

Ans.     (D)

Sol.      The point (0, –2) lies on – ve y-axis.

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37.   The distance between the points A (– 6, 7) and B (– 1, – 5) is :

(A) 13 units                                                     

(B) 14 units

(C) 15 units                                                     

(D) None of these                         

Ans.     (A)

Sol.       Distance  = \sqrt {{{( - 6 + 1)}^2} + {{(7 + 5)}^2}}

                         = 13 units

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38.   If the distance between the points (x, – 1) and (3, 2) is 5, then the value of x is :

(A) 2                                              

(B) – 2

(C) – 1                                           

(D) 1    

Ans.     (C)

Sol.      \sqrt {{{(x - 3)}^2} + {{(2 + 1)}^2}} = 5

so           x  = – 1

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39.   If the distance between points (P, – 5), (2, 7) is 13 units, then P is :

(A) – 3 or 7                                                     

(B) – 7 or 3

(C) – 3 or – 7                                                   

(D) 3 or 7                                      

Ans.     (A)

Sol.      \sqrt {{{(P - 2)}^2} + {{(7 + 5)}^2}} = 13

            on solving P  = – 3 or 7

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40.   (–5, 8) lies in which quadrant :

(A) Ist                                           

(B) 2nd

(C) 3rd                                           

(D) 4th

Ans.     (B)

Sol.      (–5, 8) lies in 2nd quadrant.

 

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