CBSE Class 10 Mathematics Coordinate Geometry Assignment Set 01

Question. Find the distance between the pairs of points : (-5,7) , (-1,3)
Answer: Let the given points be A(-5,7) and B(-1,3)
Using distance formula, we have
\(AB = \sqrt{(-1 + 5)^2 + (3 - 7)^2}\)
\(= \sqrt{4^2 + (-4)^2}\)
\(= \sqrt{16 + 16}\)
\(= \sqrt{32} = 4\sqrt{2}\) units

Question. Find the point on y-axis which is equidistant from the points (5,-2) and (-3, 2).
Answer: We know that a point on the y axis is of the form (0,y).
So, let the point P(0,y) be equidistant from A (5,-2) and B(-3,2). Then
\((5-0)^2 + (-2-y)^2 = (-3-0)^2 + (2-y)^2\)
\(25 + 4 + y^2 + 4y = 9 + 4 + y^2 - 4y\)
\(8y = -16\)
\(y = -2\)
Hence, the required point is (0,-2).
Checking :
\(AP = \sqrt{(5 - 0)^2 + (-2 + 2)^2} = \sqrt{25 + 0} = \sqrt{25} = 5\)
\(BP = \sqrt{(-3 - 0)^2 + (-2 - 2)^2} = \sqrt{9 + 16} = \sqrt{25} = 5\)

Question. Two vertices of a Triangle are (3 , -5) and ( -7, 4) . If its centroid is ( 2 , -1), find the third vertex .
Answer: Let the coordinates of the third vertex be ( x , y) . Then ,
\(\frac{x+3-7}{3}=2\) and \(\frac{y-5+4}{3}= -1\)
\(x-4 = 6\) and \(y-1= -3\)
\(x = 10\) and \(y = -2\)
Hence, third vertex of triangle is ( 10 , -2)

Question. If the mid points of the line segment joining the points P ( 6, b-2) and Q ( -2, 4) is (2, -3),find the value of b .
Answer: The coordinates of the mid-point of PQ are \([\frac{6-2}{2}, \frac{b-2+4}{2}]\)
i.e. \([2, \frac{b+2}{2}]\)
Equating it to (2 , -3)
\([\frac{b+2}{2}] = -3\)
\(b = -8\)

LEVEL – II (2 marks)

Question. The line joining the points (2,-1) and (5,-6) is bisected at P. If P lies on the line 2x + 4y + k = 0. Find the value of k.
Answer: The coordinates of P are \([\frac{2+5}{2}, \frac{-1-6}{2}]\), i.e, \(P[\frac{7}{2}, \frac{-7}{2}]\)
Since P lies on the line \(2x + 4y + k = 0\)
\(2(\frac{7}{2}) + 4(\frac{-7}{2}) + k = 0\)
\(7 - 14 + k = 0\)
\(k = 7\)

Question. 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.
Answer: Let P(x, y) divides the line segment joining A(6, 3) and B(-4, 5) in the ratio 3 :2
\(P(x,y) = P [\frac{3(-4)+ 2(6)}{3+2} , \frac{3(5)+ 2(3)}{3+2}]\)
\(= P [\frac{-12+12}{5} , \frac{15+6}{5}]\)
\(= P [0 , \frac{21}{5}]\)
Therefore, the coordinates of the point P are \((0 , \frac{21}{5})\)

Question. In each of the following find the value of ‘k’, for which the points are collinear. (7, − 2), (5, 1), (3, k)
Answer: For collinear points, area of triangle formed by them is zero
Therefore, for points (7, −2) (5, 1), and (3, k), area = 0
\(\frac{1}{2}[7\{1-k\}+5\{k-(-2)\}+3\{(-2)-1\}]=0\)
\(7-7k+5k+10-9=0\)
\(-2k+8=0\)
\(k=4\)

Question. Find the coordinates of the points of trisection of the line segment joining the points A(2 , -2) and B ( -7 , 4).
Answer: Let P and Q be the points of trisection of AB
Therefore, AP = PQ = QB
P divides AB internally in the ratio 1:2.
So, the coordinates of P , by applying the section formula are
\(\frac{1(-7) + 2(2)}{1 + 2} , \frac{1(4) + 2(-2)}{1 + 2}\)
i.e. , (-1 , 0)
now, Q also divides AB internally in the ratio 2:1 . so, the coordinates of Q are
\(\frac{2(-7) + 1(2)}{1 + 2} , \frac{2(4) + 1(-2)}{1 + 2}\)
i.e. (-4 , 2)

LEVEL – III (3 marks)

Question. If the vertices of a triangle are (1, k), (4, -3), (-9, 7) and its area is 15 sq units, find the value(s) of k.
Answer: Let A(1, k) ,B(4, -3) and C(-9, 7) be the vertices of triangle
Area of \(\Delta ABC = \frac{1}{2}[x_1 (y_2-y_3)+x_2(y_3-y_1) + x_3(y_1-y_2)]\)
\(= \frac{1}{2}[1(-3-7)+4(7-k)+(-9)(k+3)] = 15\)
\(-10 + 28 – 4k – 9k – 27 = 30\)
\(- 9 – 13k = 30\)
\(-13 k = 30+9\)
\(k = \frac{39}{-13}\)
\(k = -3\)

Question. Find the point on the x-axis which is equidistant from (2, − 5) and (− 2, 9).
Answer: We have to find a point on x-axis. Therefore, its y-coordinate will be 0.
Let the point on x-axis be (x,0)
Distance between (x,0) and \((2,-5) = \sqrt{(x-2)^2 + (0-(-5))^2} = \sqrt{(x-2)^2 + (5)^2}\)
Distance between (x,0) and \((-2,9) = \sqrt{(x-(-2))^2 + (0-(9))^2} = \sqrt{(x+2)^2 + (-9)^2}\)
By the given condition, these distances are equal in measure.
\(\sqrt{(x-2)^2 + (5)^2} = \sqrt{(x+2)^2 + (-9)^2}\)
\((x-2)^2 + 25 = (x+2)^2 + 81\)
\(x^2 - 4x + 4 + 25 = x^2 + 4x + 4 + 81\)
\(8x = 25-81\)
\(8x = -56\)
\(x = -7\)

Question. Determine the ratio in which the point P(m, 6) divides the join of A( -4, 3) and B (2,8).
Answer: Let required ratio = k:1
Using section formula \((\frac{mx_2+nx_1}{m+n} , \frac{my_2+ny_1}{m+n})\)
For y-coordinate \(6 = \frac{8k+3}{k+1}\)
\(6(k+1) = 8k + 3\)
\(6k + 6 = 8k +3\)
\(6k – 8k = 3 – 6\)
\(-2k = -3\)
\(k = 3/2\)
Therefore required ratio = 3:2

Question. Find the value of k so that the points A (-2,3), B (3,-1) and C (5,k) are collinear.
Answer: Here, \(x_1 = -2, x_2 = 3, x_3 = 5 ; y_1 = 3, y_2 = -1, y_3 = k\)
Area of \(\Delta ABC = \frac{1}{2}[ x_1(y_2-y_3) +x_2(y_3-y_1) + x_3(y_1-y_2) ]\)
\(= \frac{1}{2}[ -2(-1-k) +3(k-3) +5(3+1)]\)
\(= \frac{1}{2}[ 2 + 2k + 3k - 9 + 20]\)
\(= \frac{1}{2}[ 5k + 13 ]\)
Now, the three points will be collinear
If the area of \(\Delta ABC = 0\), i.e, if \(\frac{1}{2}[ 5k + 13 ] = 0\)
\(5k + 13 = 0\)
\(k = -\frac{13}{5}\)

LEVEL IV (4 marks)

Question. Find the value of y for which the distance between the points P(2,-3) and Q(10,y) is 10 units.
Answer: Given P(2,-3) and Q(10,y)
PQ = 10
\(PQ^2 = 10^2 = 100\)
Using distance formula
\((10-2)^2+(y-(-3))^2=100\)
\(8^2+(y+3)^2 =100\)
\(64+y^2+6y+9 =100\)
\(y^2+6y-27 = 0\)
\(y^2+9y-3y-27 = 0\)
\(y(y+9)-3(y+9) = 0\)
\((y+9)(y-3) = 0\)
\(y+9=0\) or \(y-3=0\)
Either \(y = -9\) or \(y = 3\)
Hence the required value of y can be -9 or 3

Question. If (1, 2), (4, y), (x, 6) and (3, 5) are the vertices of a parallelogram taken in order, find x and y.
Answer: Let (1, 2), (4, y), (x, 6), and (3, 5) are the coordinates of A, B, C, D vertices of a parallelogram ABCD. Intersection point O of diagonal AC and BD also divides these diagonals.
Therefore, O is the mid-point of AC and BD.
If O is the mid-point of AC, then the coordinates of O are
\(\left(\frac{1+x}{2}, \frac{2+6}{2}\right) \Rightarrow \left(\frac{x+1}{2}, 4\right)\)
If O is the mid-point of BD, then the coordinates of O are
\(\left(\frac{4+3}{2}, \frac{5+y}{2}\right) \Rightarrow \left(\frac{7}{2}, \frac{5+y}{2}\right)\)
Since both the coordinates are of the same point O,
\(\frac{x+1}{2} = \frac{7}{2}\) and \(4 = \frac{5+y}{2}\)
\(\Rightarrow x+1 = 7\) and \(5+y = 8\)
\(\Rightarrow x = 6\) and \(y = 3\)

Question. Do the points (3,2) ,(-2,-3) and (2,3) form a triangle? If so, name the type of triangle formed.
Answer: Applying the distance formula to find the distances PQ, QR, and PR, where P(3,2) , Q(-2,-3) and R(2,3) then
\(PQ = \sqrt{(-2 - 3)^2 + (-3 - 2)^2}\)
\(= \sqrt{(-5)^2 + (-5)^2} = \sqrt{25 + 25} = \sqrt{50}\)
\(QR = \sqrt{(2 + 2)^2 + (3 + 3)^2}\)
\(= \sqrt{(4)^2 + (6)^2} = \sqrt{16 + 36} = \sqrt{52}\)
\(PR = \sqrt{(2 - 3)^2 + (3 - 2)^2}\)
\(= \sqrt{(-1)^2 + (1)^2} = \sqrt{1 + 1} = \sqrt{2}\)
Since the sum of any two of these distances is greater than the third distance, the points P, Q and R form a triangle.
Also, \(PQ^2+PR^2= QR^2\)
By the converse of Pythagoras Theorem, we have \(\angle P = 90^\circ\)
Therefore, PQR is a right triangle.

Question. Do the points (3,2) ,(-2,-3) and (2,3) form a triangle? If so, name the type of triangle formed.
Answer: Applying the distance formula to find the distances PQ, QR, and PR, where P(3,2) , Q(-2,-3) and R(2,3) then
\(PQ = \sqrt{(-2 - 3)^2 + (-3 - 2)^2}\)
\(= \sqrt{(-5)^2 + (-5)^2} = \sqrt{25 + 25} = \sqrt{50}\)
\(QR = \sqrt{(2 + 2)^2 + (3 + 3)^2}\)
\(= \sqrt{(4)^2 + (6)^2} = \sqrt{16 + 36} = \sqrt{52}\)
\(PR = \sqrt{(2 - 3)^2 + (3 - 2)^2}\)
\(= \sqrt{(-1)^2 + (1)^2} = \sqrt{1 + 1} = \sqrt{2}\)
Since the sum of any two of these distances is greater than the third distance, the points P, Q and R form a triangle.
Also, \(PQ^2+PR^2= QR^2\)
By the converse of Pythagoras Theorem, we have \(\angle P = 90^\circ\)
Therefore, PQR is a right triangle.