Coordinate Geometry: Formulae and Practice Problems
Key Formulae
- Distance Formula: The distance $d$ between two points $(x_1, y_1)$ and $(x_2, y_2)$ is $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$
- Midpoint Formula: The midpoint of a line segment joining $(x_1, y_1)$ and $(x_2, y_2)$ is $\left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)$
- Section Formula (Internal): The coordinates of the point dividing the line segment joining $(x_1, y_1)$ and $(x_2, y_2)$ internally in the ratio $m:n$ are $\left(\frac{mx_2 + nx_1}{m+n}, \frac{my_2 + ny_1}{m+n}\right)$
- Centroid of a Triangle: The centroid of a triangle with vertices $(x_1, y_1)$, $(x_2, y_2)$, and $(x_3, y_3)$ is $\left(\frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3}\right)$
- Area of a Triangle: The area is given by $\frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$
- Slope ($m$) of a Line: The slope of a line passing through $(x_1, y_1)$ and $(x_2, y_2)$ is $m = \frac{y_2 - y_1}{x_2 - x_1}$
- Equation of a Line (Point-Slope Form): $y - y_1 = m(x - x_1)$
- Perpendicular Distance: The distance from a point $(x_1, y_1)$ to the line $Ax + By + C = 0$ is $d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}$
Problem 1: Distance Between Two Points
Find the distance between the points $P(2, 3)$ and $Q(5, 7)$.
Solution:
Using the distance formula: $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$
Substitute the given coordinates:
$$d = \sqrt{(5 - 2)^2 + (7 - 3)^2}$$
$$d = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5$$
The distance is $5$ units.
Problem 2: Midpoint of a Line Segment
Find the midpoint of the line segment joining the points $A(-2, 4)$ and $B(6, -8)$.
Solution:
Using the midpoint formula: $\left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)$
$x = \frac{-2 + 6}{2} = \frac{4}{2} = 2$
$y = \frac{4 + (-8)}{2} = \frac{-4}{2} = -2$
The midpoint is $(2, -2)$.
Problem 3: Section Formula
Find the coordinates of the point that divides the line segment joining the points $(1, -3)$ and $(-3, 9)$ internally in the ratio $1:3$.
Solution:
Here, $x_1=1, y_1=-3, x_2=-3, y_2=9, m=1, n=3$.
Using the section formula:
$$x = \frac{mx_2 + nx_1}{m + n} = \frac{1(-3) + 3(1)}{1 + 3} = \frac{-3 + 3}{4} = 0$$
$$y = \frac{my_2 + ny_1}{m + n} = \frac{1(9) + 3(-3)}{1 + 3} = \frac{9 - 9}{4} = 0$$
The required point is $(0, 0)$.
Problem 4: Centroid of a Triangle
Calculate the centroid of a triangle whose vertices are $(1, 2)$, $(4, -3)$, and $(-2, 7)$.
Solution:
Using the centroid formula: $\left(\frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3}\right)$
$x = \frac{1 + 4 - 2}{3} = \frac{3}{3} = 1$
$y = \frac{2 - 3 + 7}{3} = \frac{6}{3} = 2$
The centroid is $(1, 2)$.
Problem 5: Area of a Triangle
Find the area of the triangle formed by the points $A(2, 3)$, $B(-1, 0)$, and $C(2, -4)$.
Solution:
Formula: Area $= \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$
Substitute the points $(2, 3)$, $(-1, 0)$, and $(2, -4)$:
$$\text{Area} = \frac{1}{2} |2(0 - (-4)) + (-1)(-4 - 3) + 2(3 - 0)|$$
$$\text{Area} = \frac{1}{2} |2(4) - 1(-7) + 2(3)| = \frac{1}{2} |8 + 7 + 6| = \frac{1}{2} (21) = 10.5$$
The area is $10.5$ square units.
Problem 6: Finding the Slope of a Line
What is the slope of the line passing through the points $(-2, 3)$ and $(4, -5)$?
Solution:
Using the slope formula: $m = \frac{y_2 - y_1}{x_2 - x_1}$
$$m = \frac{-5 - 3}{4 - (-2)} = \frac{-8}{6} = -\frac{4}{3}$$
The slope is $-\frac{4}{3}$.
Problem 7: Equation of a Line (Point-Slope Form)
Find the equation of the line passing through the point $(3, 4)$ with a slope of $2$.
Solution:
Using the point-slope form: $y - y_1 = m(x - x_1)$
Substitute $x_1=3, y_1=4, m=2$:
$$y - 4 = 2(x - 3)$$
$$y - 4 = 2x - 6 \implies 2x - y - 2 = 0$$
The equation of the line is $2x - y - 2 = 0$.
Problem 8: Equation of a Line (Two-Point Form)
Determine the equation of the line that passes through $(-1, 1)$ and $(2, -4)$.
Solution:
First, find the slope $m$:
$$m = \frac{-4 - 1}{2 - (-1)} = \frac{-5}{3}$$
Now use the point-slope form with $(-1, 1)$:
$$y - 1 = -\frac{5}{3}(x - (-1))$$
$$3(y - 1) = -5(x + 1) \implies 3y - 3 = -5x - 5$$
$$5x + 3y + 2 = 0$$
The equation is $5x + 3y + 2 = 0$.
Problem 9: Finding Intercepts
Find the x-intercept and y-intercept of the line $3x + 4y = 12$.
Solution:
To find the x-intercept, set $y = 0$:
$3x + 4(0) = 12 \implies 3x = 12 \implies x = 4$. The x-intercept is $(4, 0)$.
To find the y-intercept, set $x = 0$:
$3(0) + 4y = 12 \implies 4y = 12 \implies y = 3$. The y-intercept is $(0, 3)$.
Problem 10: Distance from a Point to a Line
Find the perpendicular distance from the point $(2, -1)$ to the line $3x + 4y - 10 = 0$.
Solution:
Using the formula $d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}$ where $A=3, B=4, C=-10$ and the point is $(2, -1)$.
$$d = \frac{|3(2) + 4(-1) - 10|}{\sqrt{3^2 + 4^2}}$$
$$d = \frac{|6 - 4 - 10|}{\sqrt{9 + 16}} = \frac{|-8|}{5} = \frac{8}{5}$$
The perpendicular distance is $1.6$ units.
