Coordinate Geometry — CBSE Class 10 Maths Important Questions

Answer: (c) $5$5 units.

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

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

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

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

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

Answer: (a) $2 : 7$2 : 7.

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

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

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

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

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

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

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

The point is $(1, 3)$(1, 3).

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

So the ratio is $3 : 5$3 : 5.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

So the ratio is $8 : 1$8 : 1.

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

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

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

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

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

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