Sure questions of coordinate geometry

This lesson works through a set of expected Class X coordinate geometry questions. The section formula does most of the work: if $P(x, y)$ divides the segment joining $A(x_1, y_1)$ and $B(x_2, y_2)$ in the ratio $m_1 : m_2$, then

$P(x, y) = \left( \frac{m_1 x_2 + m_2 x_1}{m_1 + m_2},\ \frac{m_1 y_2 + m_2 y_1}{m_1 + m_2} \right).$

Ratio in which a point divides a segment

Find the ratio in which the point $(-4, 6)$ divides the segment joining $A(-6, 10)$ and $B(3, -8)$. Let the ratio be $k : 1$, so $m_1 = k$ and $m_2 = 1$.

The $x$-coordinate gives

$-4 = \frac{3k + (-6)}{k + 1}.$

Cross multiplying, $-4(k + 1) = 3k - 6$, so $-4k - 4 = 3k - 6$, giving $-7k = -2$ and

$k = \tfrac{2}{7}.$

So $k : 1 = 2 : 7$. Checking with the $y$-coordinate, $\dfrac{-8\left(\frac{2}{7}\right) + 10}{\frac{2}{7} + 1} = \dfrac{\frac{-16 + 70}{7}}{\frac{9}{7}} = \dfrac{54}{9} = 6$, which matches. The ratio is $2 : 7$.

Where the y-axis divides a segment

Find the ratio in which the $y$-axis divides the segment joining $A(5, -6)$ and $B(-1, -4)$, and the point of intersection. A point on the $y$-axis is $P(0, y)$. Take the ratio $k : 1$.

The $x$-coordinate is $0$:

$0 = \frac{k(-1) + 1(5)}{k + 1} \implies -k + 5 = 0 \implies k = 5.$

So the ratio is $5 : 1$. Then

$y = \frac{-4k - 6}{k + 1} = \frac{-4(5) - 6}{6} = \frac{-26}{6} = -\tfrac{13}{3}.$

The $y$-axis divides $AB$ in the ratio $5 : 1$ at the point $\left(0, -\tfrac{13}{3}\right)$.

Missing end of a diameter from the centre

Find the point $A$ where $AB$ is the diameter of a circle whose centre is $O(2, -3)$, given $B(1, 4)$. The centre is the midpoint of the diameter, so $O$ is the midpoint of $AB$. Let $A(x, y)$.

$2 = \frac{x + 1}{2} \implies x = 3, \qquad -3 = \frac{y + 4}{2} \implies y = -10.$

So $A = (3, -10)$.

A point a fraction of the way along AB

Given $A(-2, -2)$ and $B(2, -4)$, find $P$ such that $AP = \tfrac{3}{7} AB$ and $P$ lies on $AB$.

From $AP : AB = 3 : 7$, write $AP = 3x$ and $AB = 7x$, so $PB = AB - AP = 4x$. Hence

$AP : PB = 3 : 4,$

so $m_1 = 3$, $m_2 = 4$. By the section formula,

$P = \left( \frac{3(2) + 4(-2)}{7},\ \frac{3(-4) + 4(-2)}{7} \right) = \left( \frac{-2}{7},\ \frac{-20}{7} \right).$

So $P = \left(-\tfrac{2}{7}, -\tfrac{20}{7}\right)$.

Fourth vertex of a parallelogram

The points $A(6, 1)$, $B(8, 2)$, $C(9, 4)$, $D(p, 3)$ are the vertices of a parallelogram taken in order. Find $p$. The diagonals of a parallelogram bisect each other, so the midpoint of $AC$ equals the midpoint of $BD$.

$\text{mid } AC = \left( \tfrac{15}{2}, \tfrac{5}{2} \right), \qquad \text{mid } BD = \left( \tfrac{8 + p}{2}, \tfrac{5}{2} \right).$

Equating the $x$-coordinates, $\tfrac{15}{2} = \tfrac{8 + p}{2}$, so $15 = 8 + p$ and $p = 7$.

Ratio in which a line divides a segment

Determine the ratio in which the line $2x + y - 4 = 0$ divides the segment joining $A(2, -2)$ and $B(3, 7)$. Let the line meet $AB$ at $P$ in the ratio $k : 1$. By the section formula,

$P = \left( \frac{3k + 2}{k + 1},\ \frac{7k - 2}{k + 1} \right).$

Since $P$ lies on the line, it satisfies $2x + y - 4 = 0$:

$\frac{2(3k + 2) + (7k - 2) - 4(k + 1)}{k + 1} = 0.$

The numerator is $6k + 4 + 7k - 2 - 4k - 4 = 9k - 2$, so $9k - 2 = 0$ and $k = \tfrac{2}{9}$. The ratio is $2 : 9$.

Median and centroid of a triangle

The vertices of triangle $ABC$ are $A(4, 2)$, $B(6, 5)$, $C(1, 4)$. The median from $A$ meets $BC$ at $D$. Find $D$, then find the centroid $G$ with $AG : GD = 2 : 1$.

$D$ is the midpoint of $BC$:

$D = \left( \frac{6 + 1}{2},\ \frac{5 + 4}{2} \right) = \left( \tfrac{7}{2}, \tfrac{9}{2} \right).$

Now $G$ divides $AD$ in the ratio $2 : 1$, with $m_1 = 2$, $m_2 = 1$:

$G = \left( \frac{2 \cdot \frac{7}{2} + 1 \cdot 4}{3},\ \frac{2 \cdot \frac{9}{2} + 1 \cdot 2}{3} \right) = \left( \frac{11}{3}, \frac{11}{3} \right).$

Points of trisection of a segment

Find the points of trisection of the segment joining $A(2, -2)$ and $B(-7, 4)$. The points $P$ and $Q$ split $AB$ into three equal parts, so $AP : PQ : QB = 1 : 1 : 1$.

Point P divides $AB$ in the ratio $1 : 2$:

$P = \left( \frac{1(-7) + 2(2)}{3},\ \frac{1(4) + 2(-2)}{3} \right) = \left( \frac{-3}{3}, \frac{0}{3} \right) = (-1, 0).$

Point Q divides $AB$ in the ratio $2 : 1$:

$Q = \left( \frac{2(-7) + 1(2)}{3},\ \frac{2(4) + 1(-2)}{3} \right) = \left( \frac{-12}{3}, \frac{6}{3} \right) = (-4, 2).$

So the points of trisection are $(-1, 0)$ and $(-4, 2)$.

Key takeaways

• The section formula handles every case here: choosing the ratio $k : 1$ turns a division question into a single equation.
• A point on the $y$-axis has the form $(0, y)$, and the midpoint is the section formula with equal parts; the centre of a circle is the midpoint of any diameter.
• The centroid divides each median of a triangle in the ratio $2 : 1$ from the vertex.