Home Class 10 Worksheets Coordinate Geometry

Chapter 7: Coordinate Geometry

Overview

This page provides comprehensive Chapter 7: Coordinate Geometry - HOTS Worksheet - SJMaths. High Order Thinking Skills (HOTS) worksheet for Class 10 Coordinate Geometry. Advanced problems for CBSE Board Exams.

HOTS (High Order Thinking Skills)

Back to Worksheets
  1. Question 1: Find the ratio in which the line $2x + 3y - 5 = 0$ divides the line segment joining the points $(8, -9)$ and $(2, 1)$. Also find the coordinates of the point of division.
    Solution:
    Step 1: Let ratio be $k:1$. Point $P(\frac{2k+8}{k+1}, \frac{k-9}{k+1})$.
    Step 2: Substitute $P$ in line eq: $2(\frac{2k+8}{k+1}) + 3(\frac{k-9}{k+1}) - 5 = 0$.
    Step 3: $4k + 16 + 3k - 27 - 5(k+1) = 0 \Rightarrow 7k - 11 - 5k - 5 = 0 \Rightarrow 2k = 16 \Rightarrow k = 8$.
    Step 4: Ratio is 8:1. Point $P(\frac{16+8}{9}, \frac{8-9}{9}) = (8/3, -1/9)$.
  2. Question 2: If the point $P(x, y)$ is equidistant from the points $A(a+b, b-a)$ and $B(a-b, a+b)$, prove that $bx = ay$.
    Solution:
    Step 1: $PA^2 = PB^2 \Rightarrow [x-(a+b)]^2 + [y-(b-a)]^2 = [x-(a-b)]^2 + [y-(a+b)]^2$.
    Step 2: Expand and simplify: $-2x(a+b) - 2y(b-a) = -2x(a-b) - 2y(a+b)$.
    Step 3: $x(a+b) + y(b-a) = x(a-b) + y(a+b)$.
    Step 4: $ax + bx + by - ay = ax - bx + ay + by \Rightarrow 2bx = 2ay \Rightarrow bx = ay$.
  3. Question 3: The line segment joining the points $A(2, 1)$ and $B(5, -8)$ is trisected at the points $P$ and $Q$ such that $P$ is nearer to $A$. If $P$ also lies on the line given by $2x - y + k = 0$, find the value of $k$.
    Solution:
    Step 1: $P$ divides $AB$ in 1:2. $P(\frac{1(5)+2(2)}{3}, \frac{1(-8)+2(1)}{3}) = (3, -2)$.
    Step 2: $P(3, -2)$ lies on $2x - y + k = 0$.
    Step 3: $2(3) - (-2) + k = 0 \Rightarrow 6 + 2 + k = 0 \Rightarrow k = -8$.
  4. Question 4: Find the coordinates of the circumcentre of the triangle whose vertices are $(8, 6), (8, -2)$ and $(2, -2)$. Also, find its circumradius.
    Solution:
    Step 1: Let circumcentre be $O(x, y)$. $OA^2 = OB^2 = OC^2$.
    Step 2: $OA^2 = OB^2 \Rightarrow (x-8)^2 + (y-6)^2 = (x-8)^2 + (y+2)^2 \Rightarrow -12y + 36 = 4y + 4 \Rightarrow 16y = 32 \Rightarrow y = 2$.
    Step 3: $OB^2 = OC^2 \Rightarrow (x-8)^2 + (y+2)^2 = (x-2)^2 + (y+2)^2 \Rightarrow -16x + 64 = -4x + 4 \Rightarrow 12x = 60 \Rightarrow x = 5$.
    Step 4: Centre $(5, 2)$. Radius $OA = \sqrt{(5-8)^2 + (2-6)^2} = \sqrt{9+16} = 5$.
  5. Question 5: Two opposite vertices of a square are $(-1, 2)$ and $(3, 2)$. Find the coordinates of the other two vertices.
    Solution:
    Step 1: Let vertices be $A(-1, 2), C(3, 2)$. Midpoint of $AC$ is $M(1, 2)$.
    Step 2: $AC = \sqrt{(3+1)^2 + 0} = 4$. Since diagonals are equal and perpendicular, $BD = 4$ and $BD \perp AC$.
    Step 3: $AC$ is horizontal ($y=2$), so $BD$ is vertical ($x=1$).
    Step 4: Vertices $B, D$ are at distance 2 from $M(1, 2)$ on line $x=1$.
    Step 5: $B(1, 2+2) = (1, 4)$ and $D(1, 2-2) = (1, 0)$.
  6. Question 6: The midpoints of the sides of a triangle are $(3, 4), (4, 6)$ and $(5, 7)$. Find the coordinates of the vertices of the triangle.
    Solution:
    Step 1: Let vertices be $(x_1, y_1), (x_2, y_2), (x_3, y_3)$. Midpoints $D(3, 4), E(4, 6), F(5, 7)$.
    Step 2: $x_1 = x_D + x_F - x_E = 3 + 5 - 4 = 4$. $y_1 = 4 + 7 - 6 = 5$. Vertex $A(4, 5)$.
    Step 3: $x_2 = x_D + x_E - x_F = 3 + 4 - 5 = 2$. $y_2 = 4 + 6 - 7 = 3$. Vertex $B(2, 3)$.
    Step 4: $x_3 = x_E + x_F - x_D = 4 + 5 - 3 = 6$. $y_3 = 6 + 7 - 4 = 9$. Vertex $C(6, 9)$.
  7. Question 7: Prove that the points $(a, 0), (0, b)$ and $(1, 1)$ are collinear if $\frac{1}{a} + \frac{1}{b} = 1$.
    Solution:
    Step 1: Points $A(a, 0), B(0, b), C(1, 1)$. Slope $AB = \frac{b-0}{0-a} = -b/a$.
    Step 2: Slope $BC = \frac{1-b}{1-0} = 1-b$.
    Step 3: For collinearity, slopes are equal: $-b/a = 1-b \Rightarrow -b = a - ab \Rightarrow ab = a + b$.
    Step 4: Divide by $ab$: $1 = \frac{1}{b} + \frac{1}{a}$. Hence proved.
  8. Question 8: If the point $C(-1, 2)$ divides internally the line segment joining the points $A(2, 5)$ and $B(x, y)$ in the ratio $3 : 4$, find the value of $x^2 + y^2$.
    Solution:
    Step 1: $-1 = \frac{3x + 4(2)}{7} \Rightarrow -7 = 3x + 8 \Rightarrow 3x = -15 \Rightarrow x = -5$.
    Step 2: $2 = \frac{3y + 4(5)}{7} \Rightarrow 14 = 3y + 20 \Rightarrow 3y = -6 \Rightarrow y = -2$.
    Step 3: $x^2 + y^2 = (-5)^2 + (-2)^2 = 25 + 4 = 29$.
  9. Question 9: If $P(9a-2, -b)$ divides the line segment joining $A(3a+1, -3)$ and $B(8a, 5)$ in the ratio $3:1$, find the values of $a$ and $b$.
    Solution:
    Step 1: $9a-2 = \frac{3(8a) + 1(3a+1)}{4} \Rightarrow 36a - 8 = 24a + 3a + 1 \Rightarrow 9a = 9 \Rightarrow a = 1$.
    Step 2: $-b = \frac{3(5) + 1(-3)}{4} = \frac{15-3}{4} = \frac{12}{4} = 3 \Rightarrow b = -3$.
    Answer: $a=1, b=-3$.
  10. Question 10: Find the centre of a circle passing through the points $(6, -6), (3, -7)$ and $(3, 3)$.
    Solution:
    Step 1: Let centre be $O(x, y)$. $OA^2 = OB^2 \Rightarrow (x-6)^2 + (y+6)^2 = (x-3)^2 + (y+7)^2$.
    Step 2: $-12x + 36 + 12y + 36 = -6x + 9 + 14y + 49 \Rightarrow -6x - 2y + 14 = 0 \Rightarrow 3x + y = 7$.
    Step 3: $OB^2 = OC^2 \Rightarrow (x-3)^2 + (y+7)^2 = (x-3)^2 + (y-3)^2 \Rightarrow (y+7)^2 = (y-3)^2$.
    Step 4: $y+7 = \pm(y-3)$. If $+, 7=-3$ (False). If $-, y+7 = -y+3 \Rightarrow 2y = -4 \Rightarrow y = -2$.
    Step 5: $3x - 2 = 7 \Rightarrow 3x = 9 \Rightarrow x = 3$. Centre $(3, -2)$.
Previous Worksheet Next Worksheet