Coordinate Geometry — NCERT Solutions, AI Tutor & Practice

The distance formula is $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}.$

Here $(x_1, y_1) = (3, 4)$ and $(x_2, y_2) = (7, 1)$.

$d = \sqrt{(7-3)^2 + (1-4)^2} = \sqrt{4^2 + (-3)^2}$ $= \sqrt{16 + 9} = \sqrt{25} = 5.$

So the distance is $5$ units.

The origin is $O(0, 0)$. Using the distance formula: $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$ $= \sqrt{(-6 - 0)^2 + (8 - 0)^2}$ $= \sqrt{36 + 64} = \sqrt{100} = 10.$

The distance from the origin is $10$ units.

Compute all three side lengths.

$AB = \sqrt{(5-1)^2 + (2-2)^2} = \sqrt{16 + 0} = 4.$ $BC = \sqrt{(3-5)^2 + (5-2)^2} = \sqrt{4 + 9} = \sqrt{13}.$ $CA = \sqrt{(1-3)^2 + (2-5)^2} = \sqrt{4 + 9} = \sqrt{13}.$

Since $BC = CA = \sqrt{13}$, two sides are equal.

Therefore $\triangle ABC$ is isosceles.

Using the distance formula and equating to $10$: $\sqrt{(10-2)^2 + (y-(-3))^2} = 10.$ $\sqrt{64 + (y+3)^2} = 10.$

Squaring both sides: $64 + (y+3)^2 = 100.$ $(y+3)^2 = 36.$ $y + 3 = \pm 6.$

So $y + 3 = 6 \Rightarrow y = 3$, or $y + 3 = -6 \Rightarrow y = -9$.

Thus $y = 3$ or $y = -9$.

Find the three distances.

$AB = \sqrt{(4-1)^2 + (4-1)^2} = \sqrt{9+9} = \sqrt{18} = 3\sqrt{2}.$ $BC = \sqrt{(6-4)^2 + (6-4)^2} = \sqrt{4+4} = \sqrt{8} = 2\sqrt{2}.$ $AC = \sqrt{(6-1)^2 + (6-1)^2} = \sqrt{25+25} = \sqrt{50} = 5\sqrt{2}.$

Since $AB + BC = 3\sqrt{2} + 2\sqrt{2} = 5\sqrt{2} = AC$, the points are collinear.

Let the point be $P(x, 0)$ on the $x$-axis.

Equidistant means $PA = PB$, so $PA^2 = PB^2$. $(x-2)^2 + (0+5)^2 = (x+2)^2 + (0-9)^2.$ $(x-2)^2 + 25 = (x+2)^2 + 81.$ $x^2 - 4x + 4 + 25 = x^2 + 4x + 4 + 81.$ $-4x + 25 = 4x + 81.$ $-8x = 56 \Rightarrow x = -7.$

The required point is $(-7, 0)$.

Compute the squares of the side lengths.

$AB^2 = (6-3)^2 + (4-0)^2 = 9 + 16 = 25.$ $BC^2 = (-1-6)^2 + (3-4)^2 = 49 + 1 = 50.$ $CA^2 = (3-(-1))^2 + (0-3)^2 = 16 + 9 = 25.$

Isosceles: $AB^2 = CA^2 = 25$, so $AB = CA$.

Right-angled: $AB^2 + CA^2 = 25 + 25 = 50 = BC^2$, so by the converse of the Pythagoras theorem the angle at $A$ is $90^\circ$.

Hence $\triangle ABC$ is a right-angled isosceles triangle.

Set $PA^2 = PB^2$. $(k-1)^2 + (4-2)^2 = (k-3)^2 + (4+2)^2.$ $(k-1)^2 + 4 = (k-3)^2 + 36.$ $k^2 - 2k + 1 + 4 = k^2 - 6k + 9 + 36.$ $-2k + 5 = -6k + 45.$ $4k = 40 \Rightarrow k = 10.$

Thus $k = 10$.

The section formula for internal division in ratio $m_1 : m_2$ is $P = \left( \frac{m_1 x_2 + m_2 x_1}{m_1 + m_2}, \ \frac{m_1 y_2 + m_2 y_1}{m_1 + m_2} \right).$

Here $m_1 = 3$, $m_2 = 1$, $A(4,-3)$, $B(8,5)$. $x = \frac{3(8) + 1(4)}{3+1} = \frac{24 + 4}{4} = \frac{28}{4} = 7.$ $y = \frac{3(5) + 1(-3)}{3+1} = \frac{15 - 3}{4} = \frac{12}{4} = 3.$

The point is $(7, 3)$.

The midpoint formula is $M = \left( \frac{x_1 + x_2}{2}, \ \frac{y_1 + y_2}{2} \right).$

Here $P(-3, 5)$ and $Q(7, -1)$. $x = \frac{-3 + 7}{2} = \frac{4}{2} = 2.$ $y = \frac{5 + (-1)}{2} = \frac{4}{2} = 2.$

The midpoint is $(2, 2)$.

Let $P$ divide $AB$ in the ratio $k : 1$. Using the $x$-coordinate: $-4 = \frac{k(3) + 1(-6)}{k + 1}.$ $-4(k+1) = 3k - 6.$ $-4k - 4 = 3k - 6.$ $-7k = -2 \Rightarrow k = \frac{2}{7}.$

So the ratio $k : 1 = \frac{2}{7} : 1 = 2 : 7$.

Check with $y$: $\dfrac{\frac{2}{7}(-8) + 10}{\frac{2}{7}+1} = \dfrac{-\frac{16}{7} + 10}{\frac{9}{7}} = \dfrac{\frac{54}{7}}{\frac{9}{7}} = 6.$ ✓

The ratio is $2 : 7$.

In parallelogram $ABCD$, the diagonals $AC$ and $BD$ bisect each other, so their midpoints are equal.

Midpoint of $AC = \left(\dfrac{1 + x}{2}, \dfrac{2 + 6}{2}\right) = \left(\dfrac{1+x}{2}, 4\right).$

Midpoint of $BD = \left(\dfrac{4 + 3}{2}, \dfrac{y + 5}{2}\right) = \left(\dfrac{7}{2}, \dfrac{y+5}{2}\right).$

Equate the coordinates: $\frac{1+x}{2} = \frac{7}{2} \Rightarrow 1 + x = 7 \Rightarrow x = 6.$ $\frac{y+5}{2} = 4 \Rightarrow y + 5 = 8 \Rightarrow y = 3.$

Thus $x = 6$ and $y = 3$.

The two trisection points $P$ and $Q$ divide $AB$ in the ratios $1:2$ and $2:1$.

Point $P$ (ratio $1:2$): $x = \frac{1(-7) + 2(2)}{1+2} = \frac{-7 + 4}{3} = \frac{-3}{3} = -1.$ $y = \frac{1(4) + 2(-2)}{1+2} = \frac{4 - 4}{3} = 0.$ So $P(-1, 0)$.

Point $Q$ (ratio $2:1$): $x = \frac{2(-7) + 1(2)}{2+1} = \frac{-14 + 2}{3} = \frac{-12}{3} = -4.$ $y = \frac{2(4) + 1(-2)}{2+1} = \frac{8 - 2}{3} = \frac{6}{3} = 2.$ So $Q(-4, 2)$.

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

The $x$-axis has $y = 0$. Let the ratio be $k : 1$. Using the $y$-coordinate of the section formula: $0 = \frac{k(6) + 1(-3)}{k + 1}.$ $6k - 3 = 0 \Rightarrow k = \frac{1}{2}.$

So the ratio $k : 1 = \frac{1}{2} : 1 = 1 : 2$.

Point of division (using ratio $1:2$): $x = \frac{1(5) + 2(2)}{1+2} = \frac{5 + 4}{3} = \frac{9}{3} = 3.$ $y = 0 \ (\text{on the } x\text{-axis}).$

The $x$-axis divides $AB$ in ratio $1 : 2$ at the point $(3, 0)$.

Finding $a$ and $b$: Diagonals $AC$ and $BD$ have the same midpoint.

Midpoint of $AC = \left(\dfrac{-2+4}{2}, \dfrac{1+b}{2}\right) = \left(1, \dfrac{1+b}{2}\right).$

Midpoint of $BD = \left(\dfrac{a+1}{2}, \dfrac{0+2}{2}\right) = \left(\dfrac{a+1}{2}, 1\right).$

Equate: $\frac{a+1}{2} = 1 \Rightarrow a = 1, \qquad \frac{1+b}{2} = 1 \Rightarrow b = 1.$

So $B(1, 0)$ and $C(4, 1)$.

Length of $AB$: with $A(-2, 1)$ and $B(1, 0)$, $AB = \sqrt{(1-(-2))^2 + (0-1)^2} = \sqrt{9 + 1} = \sqrt{10}.$

Thus $a = 1$, $b = 1$, and $AB = \sqrt{10}$ units.

Using the section formula with $m_1 : m_2 = 2 : 3$, $A(-1, 7)$, $B(4, -3)$: $x = \frac{2(4) + 3(-1)}{2+3} = \frac{8 - 3}{5} = \frac{5}{5} = 1.$ $y = \frac{2(-3) + 3(7)}{2+3} = \frac{-6 + 21}{5} = \frac{15}{5} = 3.$

The coordinates of $P$ are $(1, 3)$.