Coordinate Geometry Test

Result

Coordinate Geometry Test - 50

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Weekly Quiz Competition

Question 1
1 / -0

Let P and Q be points $$(4, 4)$$ and $$(9, 6)$$ of parabola $${y^2} = 4a\left( {x - b} \right)$$ If R be a point on the arc of the parabola between P and Q, such that the area of $$\Delta PRQ$$ is largest, then R is

A
$$\left( {\frac{1}{4},4} \right)$$

B
$$\left( {\frac{1}{4},1} \right)$$

C
$$\left( {4,4} \right)$$

D
$$\left( {2,2\sqrt 2 } \right)$$
Question 2
1 / -0

The perpendicular bisector of the line segment joining the point A ( 1,5 ) and B ( 4,6 ) cuts the Y axis at

A
(0,13)

B
(0,-13)

C
(0,12)

D
(13,0)

Solution
Question 3
1 / -0

Area of the triangle formed by $${ (x }_{ 1 },{ y }_{ 1 })$$,$${ (x }_{ 2 },{ y }_{ 2 })$$,
$${ (3x }_{ 2 }-{ 2x }_{ 1 },{ 3y }_{ 2 }-{ 2y }_{ 1 })$$ is

A
0 sq.units

B
$${ { x }_{ 1 } }y_{ 1 } $$sq.units

C
3 sq.units

D
6 sq.units
Question 4
1 / -0

The area of a triangle whose sides are $$ a , b $$ and $$ c $$ is

A
$$ \sqrt { ( s - a ) ( s - b ) ( s - c ) } $$ sq. units

B
$$ \sqrt { s ( s + a ) ( s + b ) ( s + c ) } $$ sq. units

C
$$ \sqrt { s ( s \times a ) ( s \times b ) ( s \times c ) } $$ sq. units

D
$$ \sqrt { s ( s - a ) ( s - b ) ( s - c ) } $$ sq. units
Question 5
1 / -0

The arrangement of the areas of the triangles formed by the following points in ascending order is
i) P(0,0), Q(4,0), R(0,3)
ii) P(0,0), Q(5,0), R(0,2)
iii) P(0,0), Q(0,5), R(6,0)
iv) p(3,0), Q(0,6), R(0,0)

A
i,ii,iii,iv

B
ii,i,iii,iv

C
ii,i,iv,iii

D
iv,iii,ii,i

Solution
Question 6
1 / -0

Area of the triangle formed by co-ordinate axes and the line x + y = 5 is

A
$$\dfrac { 5 }{ 2 } $$

B
$4\dfrac { 25 }{ 2 } $$

C
$$\dfrac { 75 }{ 2 } $$

D
$$\dfrac { 125 }{ 2 } $$
Question 7
1 / -0

The centre of the circle which passes through A ($$\alpha, 0$$), B (0, $$\beta$$), C (0, 0) is

A
($$\frac{\alpha}{2},0$$)

B
($$0,\frac{\beta}{2}$$)

C
($$\frac{\alpha}{2},\frac{\beta}{2}$$)

D
($$\alpha, \beta)$$

Solution
Question 8
1 / -0

two medians drawn from the acute angles of a right angled triangle interect at the angle $$\pi /6$$ if thelength the hypotenuse of the triangle is 3 units then the area of the triangle is

A
$$\sqrt { 3 } $$

B
3

C
$$\sqrt { 2 }$$

D
9

Solution
Question 9
1 / -0

If $$m_1$$ and $$m_2$$ are the roots of the equation $$x^2 +(\sqrt 3 +2) x + \sqrt 3 - 1 = 0 $$, then the area of the triangle formed by the lines $$y=m_1x,y=-m_2x$$ and $$y=1$$ is:

A
$$\dfrac{1}{2} \Bigg ( \dfrac{-\sqrt 3 + 2}{\sqrt 3 -1} \Bigg )$$

B
$$\dfrac{1}{2} \Bigg ( \dfrac{-\sqrt 3 + 2}{\sqrt 3 +1} \Bigg )$$

C
$$\dfrac{1}{2} \Bigg ( \dfrac{\sqrt 3 + 2}{\sqrt 3 -1} \Bigg )$$

D
$$\dfrac{1}{2} \Bigg ( \dfrac{\sqrt 3 + 2}{\sqrt 3 +1} \Bigg )$$
Question 10
1 / -0

The area inside the parabola $$5x^2 - y = 0 $$ but outside the parabola $$2x^2 - y + 9 = 0$$ is

A
$$12\sqrt{3}$$ sq.units

B
$$6\sqrt{3}$$ sq.units

C
$$8\sqrt{3}$$ sq.units

D
$$4\sqrt{3}$$ sq.units