Coordinate Geometry - Test Papers

CBSE Test Paper 01

Chapter 7 Coordinate Geometry

The distance between the points $A (p s i n 25^{\circ}, 0)$ and $B (0, p s i n 65^{\circ})$ is (1)
1. 0 units
2. p units
3. p2 units
4. 1 units
If the points (x, y), (1, 2) and (7, 0) are collinear, then the relation between ‘x’ and ‘y’ is given by (1)
1. 3x – y – 7 = 0
2. 3x + y + 7 = 0
3. x + 3y – 7 = 0
4. x – 3y + 7 = 0
If the distance between the points (p, – 5) and (2, 7) is 13 units, then the value of ‘p’ is (1)
1. -3, -7
2. 3, -7
3. 3, 7
4. -3, 7
If the vertices of a triangle are (1, 1), ( – 2, 7) and (3, – 3), then its area is (1)
1. 0 sq. units
2. 2 sq. units
3. 24 sq. units
4. 12 sq. units
The distance between the points $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ is given by (1)
1. $\sqrt{{(x_{2} + x_{1})}^{2} + {(y_{2} + y_{1})}^{2}}$ units
2. $\sqrt{{(x_{2} + x_{1})}^{2} - {(y_{2} + y_{1})}^{2}}$ units
3. $\sqrt{{(x_{2} - x_{1})}^{2} - {(y_{2} - y_{1})}^{2}}$ units
4. $\sqrt{{(x_{2} - x_{1})}^{2} + {(y_{2} - y_{1})}^{2}}$ units
If the points A(x, 2), B(- 3, - 4), C(7, - 5) are collinear, then find the value of x. (1)
Find the distance between the points A and B in the following : A(a, 0), B(0, a) (1)
Find the perpendicular distance of A(5,12) from the y-axis. (1)
Find the distance of the point (- 4, - 7) from the y-axis. (1)
Find the coordinates of the centroid of a triangle whose vertices are (0,6), (8,12) and (8,0). (1)
Find the distance between the points: A(-6, -4) and B(9, -12) (2)
Find the condition that the point (x, y) may lie on the line joining (3, 4) and (-5, - 6). (2)
If P (x, y) is any point on the line joining the points A(a,0) and B(0, b), then show that $\frac{x}{a} + \frac{y}{b} = 1$ . (2)
The area of triangle formed by the points (p, 2 - 2p), (1, p, 2 p ) and (-4 -p, 6 - 2p) is 70 sq. units. How many integral values of p are possible. (3)
Point A is on x-axis, point B is on y-axis and the point P lies on line segment AB, such that P (4, - 5) and AP : PB = 5 : 3. Find the coordinates of point A and B. (3)
Show that four points (0, -1), (6, 7), (-2, 3) and (8, 3) are the vertices of a rectangle. Also, find its area. (3)
Find the co-ordinates of the points of trisection of the line segment joining the points A(1, - 2) and B(- 3,4). (3)
Show that the points A(3, 5), B(6, 0), C(1, -3) and D (-2, 2) are the vertices of a square ABCD. (4)
A (4, 2), B (6, 5) and C (1, 4) are the vertices of $△$ ABC.
1. The median from A meets BC in D. Find the coordinates of the point D.
2. Find the coordinates of point P on AD such that AP : PD = 2:1.
3. Find the coordinates of the points Q and R on medians BE and CP respectively such that BQ : QE = 2 :1 and CR: RF =2: 1.
4. What do you observe? (4)
Find the lengths of the medians of a $Δ$ ABC whose vertices are A(0, -1) B(2, 1) and C(0, 3). (4)

CBSE Test Paper 01
Chapter 07 Coordinate Geometry

Solution

1. p units
  Explanation: The distance between point A and point B=
  AB = $\sqrt{{(0 - p \sin 25^{\circ})}^{2} + {(p \sin 65^{\circ} - 0)}^{2}}$
  = $\sqrt{p^{2} \sin^{2} 25^{\circ} + p^{2} \sin^{2} 65^{\circ}}$
  = $p \sqrt{\sin^{2} 25^{\circ} + \sin^{2} (90^{\circ} - 25^{\circ})}$
  = $p \sqrt{\sin^{2} 25^{\circ} + \cos^{2} 25^{\circ}}$ $[∵ \sin (90^{\circ} - θ) = \cos θ]$
  = $p$ units
  $[∵ \cos^{2} θ + \sin^{2} θ = 1]$
1. x + 3y – 7 = 0
  Explanation: > $∴$ $\frac{1}{2} | x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2}) | = 0$
  $\Rightarrow$ $\frac{1}{2} | x (2 - 0) + 1 (0 - y) + 7 (y - 2) | = 0$
  $\Rightarrow$ $\frac{1}{2} | 2 x - y + 7 y - 14 | = 0$
  $\Rightarrow$ $2 x + 6 y - 14 = 0$ $\Rightarrow$ $x + 3 y - 7 = 0$
1. -3, 7
  Explanation: Let point A be (p, -5) and point B (2, 7) and distance between A and B = 13 units
  $∴$ $13 = \sqrt{{(2 - p)}^{2} + {(7 + 5)}^{2}}$
  $\Rightarrow$ $13 = \sqrt{4 + p^{2} - 4 p + 144}$
  $\Rightarrow$ $13 = \sqrt{p^{2} - 4 p + 148}$
  $\Rightarrow$ $169 = p^{2} - 4 p + 148$
  $\Rightarrow$ $p^{2} - 4 p - 21 = 0$
  = p2 - 7p + 3p - 21= 0
  = p(p - 7) + 3(p - 7) = 0
  $\Rightarrow$ $(p - 7) (p + 3) = 0$
  $\Rightarrow$ $p = 7, p = - 3$
1. 0 sq. units
  Explanation: Given: $(x_{1}, y_{1}) = (1, 1), (x_{2}, y_{2}) = (- 2, 7)$ and $(x_{3}, y_{3}) = (3, - 3)$ , then the Area of triangle
  $∴$ $\frac{1}{2} | x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2}) |$
  = $\frac{1}{2} | 1 (7 + 3) + (- 2) (- 3 - 1) + 3 (1 - 7) |$
  = $\frac{1}{2} | 10 + 8 - 18 |$
  = $\frac{1}{2} | 0 |$ = 0 sq. units
  Also therefore the three given points(vertices) are collinear.
1. $\sqrt{{(x_{2} - x_{1})}^{2} + {(y_{2} - y_{1})}^{2}}$ units
  Explanation: The distance between the points $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ is given by $\sqrt{(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}}$ units. This is known as distance formula.
Since the points are collinear, then,
Area of triangle = 0
$\frac{1}{2} [x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2})] = 0$
$\frac{1}{2} [x (- 4 + 5) + (- 3) (- 5 - 2) + 7 (2 + 4)] = 0$
x + 21 + 42 = 0
x = -63
A(a, 0), B(0, a)
$A B = \sqrt{{(x_{2} - x_{1})}^{2} + {(y_{2} - y_{1})}^{2}} = \sqrt{{(0 - a)}^{2} + {(a - 0)}^{2}}$
$= \sqrt{(a^{2} + a^{2})} = \sqrt{2 a^{2}} = \sqrt{2} a u n i t s$
The point on the y-axis is (0,12)
$∴$ Distance between (5,12) and (0,12)
d = $\sqrt{(0 - 5)^{2} + (12 - 12)^{2}}$
= $\sqrt{25 + 0}$
= 5 units
Points are (- 4, - 7) and (0, - 7)
Distance $= \sqrt{(0 + 4)^{2} + (- 7 + 7)^{2}}$
= $\sqrt{4^{2} + 0} = \sqrt{16}$ = 4 units
Coordinates of the centroid of a triangle whose vertices are (x1, y1), (x2, y2), (x3, y3) are $(\frac{x_{1} + x_{2} + x_{3}}{3}, \frac{y_{1} + y_{2} + y_{3}}{3})$
$= (\frac{0 + 8 + 8}{3}, \frac{6 + 12 + 0}{3})$ $= (\frac{16}{3}, \frac{18}{3}) = (\frac{16}{3}, 6)$ .
The given points are A(-6, -4) and B(9, -12)
Then, (x1 = -6, y1 = -4) and (x2 = 9, y2 = -12)
$∴ A B = \sqrt{{(x_{2} - x_{1})}^{2} + {(y_{2} - y_{1})}^{2}}$
$= \sqrt{(6 + 9)^{2} + (- 12 + 4)^{2}} = \sqrt{(15)^{2} + (- 8)^{2}}$
$= \sqrt{225 + 64} = \sqrt{289} = 17 u n i t s$
Since the point P (x, y) lies on the line joining A (3, 4) and B (-5, -6). Therefore, P, A and B are collinear points.

$∴ {4 x + 3 \times - 6 + (- 5) \times y}$ $- {3 y + (- 5) \times 4 + x \times (- 6)} = 0$
$\Rightarrow$ {4x - 18 - 5y) - (3y - 6x - 20) = 0
$\Rightarrow$ 10x -8 y + 2 = 0 $\Rightarrow$ 5x - 4 y + 1 = 0
Hence, the point (x, y) lies on the line joining (3,4) and (-5, -6), if 5x - 4y+1 = 0
It is given that the point P (x, y) lies on the line segment joining points A (a, 0) and B (0, b).
Therefore, points P (x, y), A (a, 0) and B (0, b) are collinear points.

$∴ (x \times 0 + a \times b + 0 \times y)$ $- (a \times y + 0 \times 0 + x \times b) = 0$
$\Rightarrow$ ab - (ay + bx) = 0
$\Rightarrow$ ab = ay + bx
$\Rightarrow \frac{a b}{a b} = \frac{a y}{a b} + \frac{b x}{a b}$ [Dividing throughout by ab]
$\Rightarrow 1 = \frac{y}{b} + \frac{x}{a} or \frac{x}{a} + \frac{y}{b} = 1$
Area $= \frac{1}{2}$ [p(2p - 6 + 2p) + (1 - p) (6 - 2p - 2 + 2p) + (- 4 - p)(2 - 2p - 2p)]
$\Rightarrow \frac{1}{2}$ [p(4p - 6) + (1 - p)4 + (- 4 - p) (2 - 4p)] = 70
$\Rightarrow$ 4p2 - 6p + 4 - 4p - 8 + 16p - 2p + 4p2 = 140
$\Rightarrow \frac{1}{2}$ [ -13k - 9] = 15
$\Rightarrow$ [ -13k - 9] = 30 - 13k - 9 = 30 or - 13k - 9 = - 30
k = - 3 or k $= \frac{21}{13}$
When k = - 3, coordinates = 15 sq. units
$\Rightarrow \frac{1}{2} \times A B \times$ Altitude = 15
$\Rightarrow \frac{1}{2} \times 3 \times$ Altitude = 15
$\Rightarrow$ Altitude = 10 units
Let coordinates of A are (x, 0) and coordinates of B are (0, y)

Using section formula, we get
4 = $\frac{5 \times 0 + 3 \times x}{5 + 3}$
$\Rightarrow$ 32 = 3x
$\Rightarrow$ x = $\frac{32}{3}$
Similarly, 5 = $\frac{5 \times y + 3 \times 0}{5 + 3}$
$\Rightarrow$ 40 = 5y
$\Rightarrow$ y = 8
$∴$ Coordinate of A are $(\frac{32}{3}, 0)$ and coordinates of B are (0, 8).
Let A (0 - 1), B (6, 7), C (-2, 3) and D (8, 3) be the given points. Then,
AD = $\sqrt{(8 - 0)^{2} + (3 + 1)^{2}} = \sqrt{64 + 16} = 4 \sqrt{5}$
BC = $\sqrt{(6 + 2)^{2} + (7 - 3)^{2}} = \sqrt{64 + 16} = 4 \sqrt{5}$
AC = $\sqrt{(- 2 - 0)^{2} + (3 + 1)^{2}} = \sqrt{4 + 16} = 2 \sqrt{5}$
and, BD = $\sqrt{(8 - 6)^{2} + (3 - 7)^{2}} = \sqrt{4 + 16} = 2 \sqrt{5}$
Therefore, AD = BC and AC = BD
So, ADBC is a parallelogram
Now, AB = $\sqrt{(6 - 0)^{2} + (7 + 1)^{2}} = \sqrt{36 + 64}$ = 10
and, CD = $\sqrt{(8 + 2)^{2} + (3 - 3)^{2}}$ = 10
Clearly, AB2 = AD2 + DB2 and CD2 = CB2 + BD2
Hence, ADBC is a rectangle.
Area of rectangle ADBC = $A D \times D B = (4 \sqrt{5} \times 2 \sqrt{5})$ sq. units = 40 sq. units.
Let P(x1, y1), Q(x2, y2) divides AB into 3 equal parts.
$∴$ P divides AB in the ratio of 1: 2
$∴$ $x_{1} = \frac{1 \times - 3 + 2 \times 1}{1 + 3}$ and $y_{1} = \frac{1 \times 4 + 2 \times - 2}{1 + 2}$
$\Rightarrow x_{1} = \frac{2 - 3}{3} = \frac{- 1}{3}$ $y_{1} = \frac{- 4 + 4}{3} = 0$
$∴$ Co-ordinates of P $(- \frac{1}{3}, 0)$ .
Q is the mid-point of PB.
$∴ x_{2} = \frac{\frac{- 1}{3} + (- 3)}{2}$
$= \frac{- 10}{6} = \frac{- 5}{3}$
$y_{2} = \frac{0 + 4}{2} = 2$
$∴$ Co-ordinates of Q $(- \frac{5}{3}, 2)$ .
Let A( 3,5), B(6, 0), C(1, -3 ) and D(-2, 2) be the angular points of a quadrilateral ABCD. Join AC and BD

Now $A B = \sqrt{(6 - 3)^{2} + (0 - 5)^{2}}$
$= \sqrt{3^{2} + (- 5)^{2}}$
$= \sqrt{9 + 25} = \sqrt{34} u n i t s,$
$B C = \sqrt{(1 - 6)^{2} + (- 3 - 0)^{2}} = \sqrt{(- 5)^{2} + (- 3)^{2}}$
$= \sqrt{25 + 9} = \sqrt{34} u n i t s,$
$C D = \sqrt{(- 2 - 1)^{2} + (2 + 3)^{2}} = \sqrt{(- 3)^{2} + 5^{2}}$
$= \sqrt{9 + 25} = \sqrt{34} u n i t s,$
and $D A = \sqrt{(3 + 2)^{2} + (5 - 2)^{2}} = \sqrt{5^{2} + 3^{2}}$
$= \sqrt{25 + 9} = \sqrt{34} u n i t s,$
Thus, AB = BC =CD= DA.
Diagonal $A C = \sqrt{(1 - 3)^{2} + (- 3 - 5)^{2}} = \sqrt{(- 2)^{2} + (- 8)^{2}}$
$= \sqrt{4 + 64} = \sqrt{68} = 2 \sqrt{17} u n i t s$
Diagonal $B D = \sqrt{(- 2 - 6)^{2} + (2 - 0)^{2}}$
$= \sqrt{(- 8)^{2} + 2^{2}} = \sqrt{64 + 4}$
$= \sqrt{68} = 2 \sqrt{17} u n i t s$
$∴ diag \cdot A C = diag . B D$
Thus, ABCD is a quadrilateral in which all sides are equal and the diagonals are equal.
Hence, quad. ABCD is a square.
1. Median AD of the triangle will divide the side BC in two equal parts. So D is the midpoint of side BC.
  Coordinates of $D = (\frac{6 + 1}{2}, \frac{5 + 4}{2}) = (\frac{7}{2}, \frac{9}{2})$
2. Point P divides the side AD in a ratio 2 : 1.
  Coordinates of $P = (\frac{2 \times \frac{7}{2} + 1 \times 4}{2 + 1}, \frac{2 \times \frac{9}{2} + 1 \times 2}{2 + 1})$
  $= (\frac{11}{3}, \frac{11}{3})$
3. Median BE of the triangle will divide the side AC in two equal parts. So E is the midpoint of side AC.
  Coordinates of $E = (\frac{4 + 1}{2}, \frac{2 + 4}{2}) = (\frac{5}{2}, 3)$
  Point Q divides the side BE in a ratio 2:1
  Coordinates of $Q = (\frac{2 \times \frac{5}{2} + 1 \times 6}{2 + 1}, \frac{2 \times 3 + 1 \times 5}{2 + 1})$ $= (\frac{11}{3}, \frac{11}{3})$
  Median CF of the triangle will divide the side AB in two equal parts. So F is the midpoint of side AB.
  Coordinates of $F = (\frac{4 + 6}{2}, \frac{2 + 5}{2}) = (5, \frac{7}{2})$
  Point R divides the side CF in a ratio 2:1.
  Coordinates of $R = (\frac{2 \times 5 + 1 \times 1}{2 + 1}, \frac{2 \times \frac{7}{2} + 1 \times 4}{2 + 1}) =$ $(\frac{11}{3}, \frac{11}{3})$
4. Now we may observe that coordinates of point P, Q are same. So, all these are representing same point on the plane i.e. centroid of the triangle.
Let D, E, F be the midpoint of the side BC, CA and AB respectively in $Δ$ ABC

Then, by the midpoint formula, we have
$D (\frac{2 + 0}{2}, \frac{1 + 3}{2}), E (\frac{0 + 0}{2}, \frac{3 - 1}{2})$ $F (\frac{0 + 2}{2}, \frac{- 1 + 1}{2})$
i.e., D(1, 2), E(0, 1), F(1, 0)
Hence the lengths of medians AD, BE and CF are given by
$A D = \sqrt{(1 - 0)^{2} + (2 + 1)^{2}} = \sqrt{1 + 9} = \sqrt{10} u n i t s$
$B E = \sqrt{(0 - 2)^{2} + (1 - 1)^{2}} = \sqrt{4 + 0} = \sqrt{4} = 2 u n i t s$
$C F = \sqrt{(1 - 0)^{2} + (0 - 3)^{2}} = \sqrt{1 + 9} = \sqrt{10} u n i t s$
Hence, AD = $\sqrt{10},$ BE= 2, CF = $\sqrt{10}$