Math II

Points and lines (coordinate geometry): practice problems

Coordinates let you study geometry by calculation. Make the distance, section and line formulas second nature.

Basic, Standard, Advanced: Grade 11 · Term 1

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Distance and section formulas

Formulas

For $\mathrm{A}(x_1,\ y_1)$ and $\mathrm{B}(x_2,\ y_2)$ :

\mathrm{AB} = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}

The point dividing AB internally in the ratio $m : n$ is $\left(\frac{nx_1 + mx_2}{m + n},\ \frac{ny_1 + my_2}{m + n}\right)$ (replace $n$ by $-n$ for external division).

The centroid of a triangle is $\left(\frac{x_1 + x_2 + x_3}{3},\ \frac{y_1 + y_2 + y_3}{3}\right)$ .

Equations of lines; parallel and perpendicular

The line through $(x_1,\ y_1)$ with slope $m$ is $y - y_1 = m(x - x_1)$ . For a line through two points, find the slope first.

Parallel and perpendicular

For $y = m_1x + n_1$ and $y = m_2x + n_2$ : parallel when $m_1 = m_2$ , perpendicular when $m_1m_2 = -1$ .

For $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$ : parallel when $a_1b_2 - a_2b_1 = 0$ , perpendicular when $a_1a_2 + b_1b_2 = 0$ .

A line such as $k(x + 2y - 3) + (x - y) = 0$ passes through the intersection of $x + 2y - 3 = 0$ and $x - y = 0$ for every $k$ .

Point-to-line distance and reflections

Distance from a point to a line

d = \frac{|ax_1 + by_1 + c|}{\sqrt{a^2 + b^2}}

To reflect A in a line $\ell$ , let the image be $(p,\ q)$ and use two conditions: the midpoint lies on $\ell$ , and the segment is perpendicular to $\ell$ .

To minimise $\mathrm{AP} + \mathrm{PB}$ for P on a line, reflect A to A′: then $\mathrm{AP} + \mathrm{PB} = \mathrm{A'P} + \mathrm{PB} \geq \mathrm{A'B}$ .

Q1BasicGr 11 · Term 1Points and lines

Find the distance between the points $\mathrm{A}(-5,\ 1)$ and $\mathrm{B}(-8,\ -3)$ .

Answer

\mathrm{AB} = 5

Solution

By the distance formula,
$\mathrm{AB} = \sqrt{\{-8 - (-5)\}^2 + (-3 - 1)^2} = \sqrt{(-3)^2 + (-4)^2} = \sqrt{25} = 5$

Q2StandardGr 11 · Term 1Points and lines

Find the distance between the point $\mathrm{A}(4,\ 0)$ and the line $4x + 3y - 7 = 0$ .

Answer

\frac{9}{5}

Solution

By the point-to-line distance formula,
$d = \frac{|4 \cdot 4 + 3 \cdot 0 - 7|}{5} = \frac{9}{5}$

Q3AdvancedGr 11 · Term 1Points and lines

Find the coordinates of the circumcentre of $\triangle \mathrm{ABC}$ with vertices $\mathrm{A}(1,\ 0)$ , $\mathrm{B}(-3,\ -2)$ and $\mathrm{C}(-3,\ -4)$ .

Answer

(0,\ -3)

Solution

Let the circumcentre be P $(x,\ y)$ . From $\mathrm{PA}^2 = \mathrm{PB}^2$ ,
$2x + y = -3$
From $\mathrm{PA}^2 = \mathrm{PC}^2$ ,
$x + y = -3$
Solving,
$x = 0,\quad y = -3$

Q1BasicGr 11 · Term 1Points and lines

Find the equation of the line through $\mathrm{A}(-3,\ -5)$ and $\mathrm{B}(-5,\ 5)$ .

Q2StandardGr 11 · Term 1Points and lines

Find the value(s) of the constant $k$ for which the lines $(k + 1)x + 12y + 4 = 0$ and $x + ky - 2 = 0$ satisfy each condition.

(1)They are parallel.
(2)They are perpendicular.

Q3AdvancedGr 11 · Term 1Points and lines

The line $(k + 2)x + (-2k - 2)y + k + 2 = 0$ passes through a fixed point for every value of the constant $k$ . Find the coordinates of that point.

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Points and lines (coordinate geometry): practice problems

Distance and section formulas

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