Grade X Topic: Coordinate Geometry Level - 1 Worksheets for ... - Free Printable

Here are the step-by-step solutions for the problems in your worksheet.

Multiple Choice Questions (MCQs)

1. Find the coordinates of the point equidistant from the points A(1, 2), B (3, -4) and C(5, -6).
* Step 1: Let the required point be $P(x, y)$. Since it is equidistant, the distance $PA = PB = PC$. It is easier to work with squared distances: $PA^2 = PB^2 = PC^2$.
* Step 2: Set $PA^2 = PB^2$:
$(x-1)^2 + (y-2)^2 = (x-3)^2 + (y+4)^2$
$x^2 - 2x + 1 + y^2 - 4y + 4 = x^2 - 6x + 9 + y^2 + 8y + 16$
Cancel $x^2$ and $y^2$, then simplify:
$-2x - 4y + 5 = -6x + 8y + 25$
$4x - 12y = 20 \Rightarrow x - 3y = 5$ (Equation 1)
* Step 3: Set $PB^2 = PC^2$:
$(x-3)^2 + (y+4)^2 = (x-5)^2 + (y+6)^2$
$x^2 - 6x + 9 + y^2 + 8y + 16 = x^2 - 10x + 25 + y^2 + 12y + 36$
Simplify:
$-6x + 8y + 25 = -10x + 12y + 61$
$4x - 4y = 36 \Rightarrow x - y = 9$ (Equation 2)
* Step 4: Solve the system. From Eq 2, $x = y + 9$. Substitute into Eq 1:
$(y + 9) - 3y = 5$
$-2y = -4 \Rightarrow y = 2$
$x = 2 + 9 = 11$
* Result: The point is (11, 2).

2. Find the distance of the point (-6, 8) from the origin.
* Formula: Distance from origin $(0,0)$ is $\sqrt{x^2 + y^2}$.
* Calculation: $\sqrt{(-6)^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10$.

3. Find the ratio in which the line joining the points (6, 4) and (1, -7) is divided by x-axis.
* Concept: Any point on the x-axis has a y-coordinate of 0.
* Formula: Section formula for y-coordinate: $y = \frac{m y_2 + n y_1}{m + n}$.
* Calculation: Let ratio be $k:1$.
$0 = \frac{k(-7) + 1(4)}{k + 1}$
$0 = -7k + 4 \Rightarrow 7k = 4 \Rightarrow k = \frac{4}{7}$.
* Result: The ratio is 4 : 7.

4. Mid-point of the line-segment joining the points (-5, 4) and (9, -8) is:
* Formula: Midpoint $= (\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2})$.
* Calculation:
x-coord: $\frac{-5 + 9}{2} = \frac{4}{2} = 2$
y-coord: $\frac{4 + (-8)}{2} = \frac{-4}{2} = -2$
* Result: The midpoint is (2, -2).

5. The distance of the point P(2, 3) from the x-axis is
* Concept: The perpendicular distance of a point $(x,y)$ from the x-axis is simply the absolute value of its y-coordinate ($|y|$).
* Calculation: Point is $(2, 3)$. The y-coordinate is 3. Distance is 3.

6. The points (1,1), (-2, 7) and (3, -3) are
* Method: Calculate slopes between pairs. If slopes are equal, they are collinear.
Slope AB: $\frac{7-1}{-2-1} = \frac{6}{-3} = -2$
Slope BC: $\frac{-3-7}{3-(-2)} = \frac{-10}{5} = -2$
* Result: Since slopes are equal, the points are collinear.

---

Answer the following questions

7. Point A (1, 2) and B (3, 4) are two ends of a line segment. Find the point which divides AB in the ratio 3:4.
* Formula: Section Formula: $(\frac{mx_2+nx_1}{m+n}, \frac{my_2+ny_1}{m+n})$ where $m=3, n=4$.
* x-coordinate: $\frac{3(3) + 4(1)}{3+4} = \frac{9+4}{7} = \frac{13}{7}$
* y-coordinate: $\frac{3(4) + 4(2)}{3+4} = \frac{12+8}{7} = \frac{20}{7}$
* Answer: $(\frac{13}{7}, \frac{20}{7})$

8. If (1, 2), (4, y), (x, 6) and (3, 5) are the vertices of a parallelogram taken in order, find x and y.
* Property: Diagonals of a parallelogram bisect each other. This means the midpoint of diagonal AC is the same as the midpoint of diagonal BD.
Vertices: $A(1,2), B(4,y), C(x,6), D(3,5)$.
* Midpoint of AC: $(\frac{1+x}{2}, \frac{2+6}{2}) = (\frac{1+x}{2}, 4)$
* Midpoint of BD: $(\frac{4+3}{2}, \frac{y+5}{2}) = (\frac{7}{2}, \frac{y+5}{2})$
* Equating Coordinates:
x-coords: $\frac{1+x}{2} = \frac{7}{2} \Rightarrow 1+x=7 \Rightarrow x=6$
y-coords: $4 = \frac{y+5}{2} \Rightarrow 8=y+5 \Rightarrow y=3$
* Answer: $x = 6, y = 3$

9. Find a point on the y-axis equidistant from {-5, 2} and {9, -2}.
* Setup: Let the point on the y-axis be $P(0, y)$. Let $A=(-5, 2)$ and $B=(9, -2)$.
* Condition: $PA^2 = PB^2$
$(0 - (-5))^2 + (y - 2)^2 = (0 - 9)^2 + (y - (-2))^2$
$25 + (y^2 - 4y + 4) = 81 + (y^2 + 4y + 4)$
$y^2 - 4y + 29 = y^2 + 4y + 85$
* Solve: Subtract $y^2$ from both sides:
$-4y + 29 = 4y + 85$
$29 - 85 = 8y$
$-56 = 8y \Rightarrow y = -7$
* Answer: The point is (0, -7).

10. Find the distance between the points (-8/5, 2) and (2/5, 2).
* Observation: The y-coordinates are the same (both are 2). This is a horizontal line.
* Calculation: Distance is simply the difference in x-coordinates.
Distance $= | \frac{2}{5} - (\frac{-8}{5}) | = | \frac{2}{5} + \frac{8}{5} | = \frac{10}{5} = 2$.
* Answer: 2 units.

──────────────────────────────────────

Final Answer:
MCQs:
1. (d) (11, 2)
2. (c) 10
3. (c) 4 : 7
4. (c) (2, -2)
5. (b) 3
6. (b) collinear

Short Answers:
7. $(\frac{13}{7}, \frac{20}{7})$
8. $x = 6, y = 3$
9. $(0, -7)$
10. 2