Coordinate Geometry Sure Questions Part 2 (Class 10)

This lesson works through a set of sure-fire Class 10 coordinate geometry questions: distances between points, testing for triangles and right triangles, collinearity, equidistant points, the section formula, and midpoints.

Distance between two points

The distance between $A(x_1, y_1)$ and $B(x_2, y_2)$ is

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

For $A(0,5)$ and $B(0,8)$:

$AB = \sqrt{(0-0)^2 + (5-8)^2} = \sqrt{0 + 9} = 3.$

When the $x$-coordinates are equal, the distance is just the gap in the $y$-coordinates, $|y_2 - y_1| = |8 - 5| = 3$. Likewise, when the $y$-coordinates are equal, the distance is $|x_2 - x_1|$.

The distance of a point $(x, y)$ from the origin is $\sqrt{x^2 + y^2}$. For $(3,4)$:

$\sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5.$

Do three points form a triangle?

Take $A(3,2)$, $B(-2,-3)$, $C(2,3)$. Find the three side lengths.

$AB = \sqrt{(3-(-2))^2 + (2-(-3))^2} = \sqrt{5^2 + 5^2} = \sqrt{50} \approx 7.07.$

$BC = \sqrt{(-2-2)^2 + (-3-3)^2} = \sqrt{(-4)^2 + (-6)^2} = \sqrt{52} \approx 7.21.$

$AC = \sqrt{(3-2)^2 + (2-3)^2} = \sqrt{1^2 + (-1)^2} = \sqrt{2} \approx 1.41.$

Three points form a triangle if the sum of any two sides is greater than the third. Here every pair satisfies this, so $A$, $B$, $C$ form a triangle.

Is it a right triangle?

The longest side is $BC = \sqrt{52}$, so $BC^2 = 52$. Check the other two sides:

$AB^2 + AC^2 = (\sqrt{50})^2 + (\sqrt{2})^2 = 50 + 2 = 52 = BC^2.$

Since the square of the longest side equals the sum of the squares of the other two, $ABC$ is a right triangle.

Proving points are collinear

Take $A(3,1)$, $B(6,4)$, $C(8,6)$.

$AB = \sqrt{(3-6)^2 + (1-4)^2} = \sqrt{9 + 9} = \sqrt{18} = 3\sqrt{2}.$

$BC = \sqrt{(6-8)^2 + (4-6)^2} = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2}.$

$AC = \sqrt{(3-8)^2 + (1-6)^2} = \sqrt{25 + 25} = \sqrt{50} = 5\sqrt{2}.$

Since $AB + BC = 3\sqrt{2} + 2\sqrt{2} = 5\sqrt{2} = AC$, the three points lie on a straight line, so $A$, $B$, $C$ are collinear.

Point on the y-axis equidistant from two points

Find the point on the $y$-axis equidistant from $A(2,-5)$ and $B(-2,9)$. A point on the $y$-axis has the form $P(0, y)$, and $PA = PB$:

$\sqrt{(2-0)^2 + (-5-y)^2} = \sqrt{(-2-0)^2 + (9-y)^2}.$

Squaring both sides:

$4 + 25 + 10y + y^2 = 4 + 81 - 18y + y^2.$

The $4$ and $y^2$ cancel, leaving $25 + 10y = 81 - 18y$, so $28y = 56$ and $y = 2$. The point is $(0, 2)$.

Finding y from a fixed distance

Find $y$ so that the distance between $P(2,-3)$ and $Q(10, y)$ is $10$.

$\sqrt{(2-10)^2 + (-3-y)^2} = 10.$

Squaring: $64 + (y^2 + 6y + 9) = 100$, so

$y^2 + 6y - 27 = 0.$

Factorising (product $-27$, sum $6$ gives $9$ and $-3$): $(y+9)(y-3) = 0$, so $y = -9$ or $y = 3$.

A relation between x and y

Find the relation between $x$ and $y$ so that $P(x,y)$ is equidistant from $A(7,1)$ and $B(3,5)$. From $PA = PB$, squaring both sides:

$(x-7)^2 + (y-1)^2 = (x-3)^2 + (y-5)^2.$

Expanding:

$x^2 - 14x + 49 + y^2 - 2y + 1 = x^2 - 6x + 9 + y^2 - 10y + 25.$

The $x^2$ and $y^2$ terms cancel, leaving $-14x - 2y + 50 = -6x - 10y + 34$. Collecting terms:

$-8x + 8y = -16 \quad\Longrightarrow\quad x - y = 2.$

Section formula

The point $P(x,y)$ that divides the segment from $A(x_1,y_1)$ to $B(x_2,y_2)$ internally in the ratio $m_1 : m_2$ is

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

For $A(4,-3)$ and $B(8,5)$ divided in the ratio $3 : 1$ (so $m_1 = 3$, $m_2 = 1$):

$P = \left( \frac{3 \cdot 8 + 1 \cdot 4}{4}, \; \frac{3 \cdot 5 + 1 \cdot (-3)}{4} \right) = \left( \frac{28}{4}, \frac{12}{4} \right) = (7, 3).$

Midpoint of a segment

The midpoint of $P(x_1,y_1)$ and $Q(x_2,y_2)$ is

$\left( \frac{x_1 + x_2}{2}, \; \frac{y_1 + y_2}{2} \right).$

For $P(3,9)$ and $Q(-3,-7)$:

$\left( \frac{3 + (-3)}{2}, \; \frac{9 + (-7)}{2} \right) = (0, 1).$

Key takeaways

• The distance formula $\sqrt{(x_1-x_2)^2 + (y_1-y_2)^2}$ underlies triangle, collinearity, and equidistant-point questions.
• Three points form a triangle when each pair of sides beats the third; they are collinear when two side lengths add up to the third.
• A triangle is right-angled when the square of its longest side equals the sum of the squares of the other two.
• The section formula divides a segment in a given ratio, and the midpoint formula is the special case of the ratio $1:1$.