Coordinate Geometry and Algebraic Exercises

Vocabulary and key algebraic steps from the January 2016 lecture notes covering coordinate geometry distance equations Ex-6 and Ex-8.

Ex-6

A coordinate distance problem based on the equation $\sqrt{(4-2)^2 - (3-3)^2} = \sqrt{(2-x)^2 + (3-5)^2}$, which simplifies to find $x = 2$.

Quadratic Equation in Ex-6

The equation $x^2 - 4x + 4 = 0$, formed by setting $x^2 - 4x + 8 = 4$ and subtracting $4$ from both sides.

Factorization in Ex-6

The factoring process rewriting $x^2 - 4x + 4 = 0$ as $x^2 - 2x - 2x + 4 = 0$, grouped as $x(x-2) - 2(x-2) = 0$, yielding $(x-2)(x-2) = 0$.

Root of Ex-6

The final value solved from the repeated factors $(x-2)(x-2) = 0$, which is $x = 2$.

Point Representation in Ex-8

The coordinates defined by the relation $x = 2y$, written as $(2y, y)$.

Radical Equation in Ex-8

The initial distance equality $\sqrt{(2y-2)^2 + (y+5)^2} = \sqrt{(2y+3)^2 + (y-6)^2}$.

LHS Expansion in Ex-8

The expansion of $(2y-2)^2 + (y+5)^2$, which yields $4y^2 + 4 - 8y + y^2 + 25 + 10y$.

RHS Expansion in Ex-8

The expansion of $(2y+3)^2 + (y-6)^2$, which yields $4y^2 + 9 + 12y + y^2 + 36 - 12y$.

Ex-8 Solution for $y$

The final variable solution obtained from $2y = 45 + 29$, calculated as $y = \frac{74}{2} = 37$.