Coordinate Geometry — Important Questions

Answer: (c) $5$ units.

Distance from the origin $= \sqrt{x^2 + y^2} = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5$ units.

Answer: (a) $(3, 5)$.

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

Answer: (a) $(-7, 0)$.

Let the point be $P(x, 0)$. Then $PA^2 = PB^2$: $(x - 2)^2 + 25 = (x + 2)^2 + 81$. Expanding: $-4x + 25 = 4x + 81 \Rightarrow -8x = 56 \Rightarrow x = -7$.

Answer: (a) $2 : 7$.

Let the ratio be $k : 1$. Using the $x$-coordinate: $-4 = \dfrac{3k - 6}{k + 1} \Rightarrow -4k - 4 = 3k - 6 \Rightarrow 7k = 2 \Rightarrow k = \dfrac{2}{7}$. Hence the ratio is $2 : 7$.

Answer: (a) Both A and R are true and R is the correct explanation of A.

Every point on the $y$-axis has $x$-coordinate $0$. Since $(0, -3)$ has $x$-coordinate $0$, it lies on the $y$-axis, exactly as R explains.

$PQ^2 = (10 - 2)^2 + (y + 3)^2 = 100 \Rightarrow 64 + (y + 3)^2 = 100 \Rightarrow (y + 3)^2 = 36$.

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

Using the section formula with $m_1 : m_2 = 2 : 3$:

$x = \dfrac{2(4) + 3(-1)}{2 + 3} = \dfrac{8 - 3}{5} = 1$, $\quad y = \dfrac{2(-3) + 3(7)}{5} = \dfrac{-6 + 21}{5} = 3$.

The point is $(1, 3)$.

On the $x$-axis, $y = 0$. Let the ratio be $k : 1$. Using the $y$-coordinate: $0 = \dfrac{5k - 3}{k + 1} \Rightarrow 5k = 3 \Rightarrow k = \dfrac{3}{5}$.

So the ratio is $3 : 5$.

$x = \dfrac{3(4) + 5(1)}{3 + 5} = \dfrac{12 + 5}{8} = \dfrac{17}{8}$.

The point of division is $\left(\dfrac{17}{8},\ 0\right)$.

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

$P$ (ratio $1 : 2$): $x = \dfrac{1(-2) + 2(4)}{3} = \dfrac{6}{3} = 2$, $\ y = \dfrac{1(-3) + 2(-1)}{3} = \dfrac{-5}{3}$. So $P\left(2, -\dfrac{5}{3}\right)$.

$Q$ (ratio $2 : 1$): $x = \dfrac{2(-2) + 1(4)}{3} = 0$, $\ y = \dfrac{2(-3) + 1(-1)}{3} = \dfrac{-7}{3}$. So $Q\left(0, -\dfrac{7}{3}\right)$.

$AB = \sqrt{(4 - 1)^2 + (2 - 7)^2} = \sqrt{9 + 25} = \sqrt{34}$.

$BC = \sqrt{(-1 - 4)^2 + (-1 - 2)^2} = \sqrt{25 + 9} = \sqrt{34}$.

$CD = \sqrt{(-4 + 1)^2 + (4 + 1)^2} = \sqrt{9 + 25} = \sqrt{34}$.

$DA = \sqrt{(1 + 4)^2 + (7 - 4)^2} = \sqrt{25 + 9} = \sqrt{34}$.

All four sides are equal. The diagonals: $AC = \sqrt{(-1 - 1)^2 + (-1 - 7)^2} = \sqrt{4 + 64} = \sqrt{68}$ and $BD = \sqrt{(-4 - 4)^2 + (4 - 2)^2} = \sqrt{64 + 4} = \sqrt{68}$.

Since all four sides are equal and the two diagonals are equal, $ABCD$ is a square.

In a parallelogram the diagonals bisect each other, so the midpoint of $AC$ equals the midpoint of $BD$.

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

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

Equating coordinates: $\dfrac{a + 1}{2} = 1 \Rightarrow a = 1$; and $\dfrac{1 + b}{2} = 1 \Rightarrow b = 1$.

So the vertices are $A(-2, 1)$, $B(1, 0)$, $C(4, 1)$, $D(1, 2)$.

$AB = \sqrt{(1 + 2)^2 + (0 - 1)^2} = \sqrt{9 + 1} = \sqrt{10}$ units.

$BC = \sqrt{(4 - 1)^2 + (1 - 0)^2} = \sqrt{9 + 1} = \sqrt{10}$ units.

Let the line divide the segment in the ratio $k : 1$ at a point $P$. By the section formula:

$P = \left(\dfrac{2k + 8}{k + 1},\ \dfrac{k - 9}{k + 1}\right)$.

Since $P$ lies on $2x + 3y - 5 = 0$:

$2\left(\dfrac{2k + 8}{k + 1}\right) + 3\left(\dfrac{k - 9}{k + 1}\right) - 5 = 0$

$\Rightarrow (4k + 16) + (3k - 27) - 5(k + 1) = 0 \Rightarrow 2k - 16 = 0 \Rightarrow k = 8$.

So the ratio is $8 : 1$.

$x = \dfrac{2(8) + 8}{8 + 1} = \dfrac{24}{9} = \dfrac{8}{3}$, $\quad y = \dfrac{8 - 9}{9} = -\dfrac{1}{9}$.

The point of division is $\left(\dfrac{8}{3},\ -\dfrac{1}{9}\right)$.

(i) $AB = \sqrt{(6 - 3)^2 + (7 - 4)^2} = \sqrt{9 + 9} = \sqrt{18} = 3\sqrt{2}$ m.

(ii) $T = \left(\dfrac{3 + 9}{2}, \dfrac{4 + 4}{2}\right) = (6, 4)$.

(iii) $BT = \sqrt{(6 - 6)^2 + (7 - 4)^2} = \sqrt{0 + 9} = 3$ m.

OR Point dividing $AB$ in the ratio $1 : 2$: $\left(\dfrac{1(6) + 2(3)}{3}, \dfrac{1(7) + 2(4)}{3}\right) = \left(\dfrac{12}{3}, \dfrac{15}{3}\right) = (4, 5)$.