Coordinate geometry, also known as analytic geometry, is the study of geometry using a coordinate system. This integration of algebra and geometry allows us to solve geometric problems using algebraic equations. Below are the essential formulas and 8 key problems with detailed solutions to help you master the basics.
Key Formulas Cheat Sheet
Solved Problems
Find the distance between the points $A(2, 3)$ and $B(5, 7)$.
Solution:
Let $(x_1, y_1) = (2, 3)$ and $(x_2, y_2) = (5, 7)$.
Using the distance formula:
$$d = \sqrt{(5 - 2)^2 + (7 - 3)^2}$$ $$d = \sqrt{(3)^2 + (4)^2}$$ $$d = \sqrt{9 + 16}$$ $$d = \sqrt{25}$$Find the coordinates of the midpoint of the line segment joining the points $P(-4, 2)$ and $Q(8, 6)$.
Solution:
Using the midpoint formula $M = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right)$:
$$x = \frac{-4 + 8}{2} = \frac{4}{2} = 2$$ $$y = \frac{2 + 6}{2} = \frac{8}{2} = 4$$Find the coordinates of the point which divides the line segment joining the points $(4, -3)$ and $(8, 5)$ in the ratio $3:1$ internally.
Solution:
Here, $x_1=4, y_1=-3, x_2=8, y_2=5$, and ratio $m_1:m_2 = 3:1$.
Using the Section Formula:
$$x = \frac{3(8) + 1(4)}{3 + 1} = \frac{24 + 4}{4} = \frac{28}{4} = 7$$ $$y = \frac{3(5) + 1(-3)}{3 + 1} = \frac{15 - 3}{4} = \frac{12}{4} = 3$$Determine if the points $A(1, 5)$, $B(2, 3)$, and $C(-2, 11)$ are collinear.
Solution:
Points are collinear if the area of the triangle formed by them is zero.
$$Area = \frac{1}{2} |1(3 - 11) + 2(11 - 5) + (-2)(5 - 3)|$$ $$Area = \frac{1}{2} |1(-8) + 2(6) - 2(2)|$$ $$Area = \frac{1}{2} |-8 + 12 - 4|$$ $$Area = \frac{1}{2} |0| = 0$$Find the value of $x$ if the point $(x, 2)$ is equidistant from $(8, -2)$ and $(2, -2)$.
Solution:
Let $P(x, 2)$, $A(8, -2)$, and $B(2, -2)$. We are given $PA = PB$, so $PA^2 = PB^2$.
$$(x - 8)^2 + (2 - (-2))^2 = (x - 2)^2 + (2 - (-2))^2$$ $$(x - 8)^2 + 4^2 = (x - 2)^2 + 4^2$$Subtract $16$ ($4^2$) from both sides:
$$(x - 8)^2 = (x - 2)^2$$ $$x^2 - 16x + 64 = x^2 - 4x + 4$$Cancel $x^2$ and solve for $x$:
$$-16x + 4x = 4 - 64$$ $$-12x = -60$$ $$x = 5$$Find the centroid of a triangle whose vertices are $(3, -5)$, $(-7, 4)$, and $(10, -2)$.
Solution:
The formula for the centroid is $G(x, y) = \left( \frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3} \right)$.
$$x = \frac{3 + (-7) + 10}{3} = \frac{6}{3} = 2$$ $$y = \frac{-5 + 4 + (-2)}{3} = \frac{-3}{3} = -1$$Calculate the area of the triangle formed by the vertices $A(2, 3)$, $B(-1, 0)$, and $C(2, -4)$.
Solution:
Using the area formula:
$$Area = \frac{1}{2} |2(0 - (-4)) + (-1)(-4 - 3) + 2(3 - 0)|$$ $$Area = \frac{1}{2} |2(4) - 1(-7) + 2(3)|$$ $$Area = \frac{1}{2} |8 + 7 + 6|$$ $$Area = \frac{1}{2} |21|$$Find the ratio in which the y-axis divides the line segment joining the points $(5, -6)$ and $(-1, -4)$.
Solution:
Any point on the y-axis has coordinates $(0, y)$. Let the ratio be $k:1$.
Using the section formula for the x-coordinate (which we know is 0):
$$x = \frac{m_1x_2 + m_2x_1}{m_1 + m_2}$$ $$0 = \frac{k(-1) + 1(5)}{k + 1}$$Cross multiply:
$$0 = -k + 5$$ $$k = 5$$