Question 1

In the given diagram, ΔAOB is a right-angled triangle and C is the mid-point of AB. The coordinates of the point which is equidistant from the three vertices of ΔAOB is:

Accepted Answer

Correct Option: (a) (x, y) Theorem In a right-angled triangle, the circumcentre (equidistant point from all 3 vertices) is the midpoint of the hypotenuse . Coordinates of A = (0, 2y), B = (2x, 0). Midpoint C = [ (0 + 2x)/2, (2y + 0)/2 ] = (x, y) .

Question 2

Line segment joining A(1, −3) and B(−4, 5) is divided by the y-axis. Find the: (a) ratio in which the y-axis divides AB. (b) coordinates of the point of intersection.

Accepted Answer

Part (a) Any point on the y-axis has x-coordinate = 0. Let the ratio be k : 1.
By section formula: x = [k(−4) + 1(1)] / (k + 1) = 0
⇒ −4k + 1 = 0 ⇒ 4k = 1 ⇒ k = 1/4.
Hence, the ratio is 1 : 4. [1.5 Marks]

Part (b) y-coordinate = [1(5) + 4(−3)] / (1 + 4) = (5 − 12) / 5 = −7/5.
Coordinates of point of intersection = (0, −7/5) or (0, −1.4). [1.5 Marks]

Question 3

In the parallelogram ABCD with vertices A(3, 3), B(0, −3), C(−4, −3), D(−1, 3), calculate the coordinates of 'P', the point of intersection of diagonals AC and BD.

Accepted Answer

Diagonals of a parallelogram bisect each other. P is midpoint of AC: x = (3 + (−4)) / 2 = −1/2 = −0.5; y = (3 + (−3)) / 2 = 0. Coordinates of P = (−0.5, 0) . [1 Mark]

Question 4

In the given diagram, ABC is a triangle, where B (4, −4) and C (−4, −2). D is a point on AC . (a) Write down the coordinates of A and D . (b) Find the coordinates of the centroid of Δ ABC . (c) If D divides AC in the ratio k : 1, find the value of k . (d) Find the equation of the line BD .

Accepted Answer

(a) From graph: A (0, 6) , D (−3, 0) . [1 Mark] (b) Centroid G = ((0+4−4)/3, (6−4−2)/3) = (0, 0) . [1 Mark] (c) x D = (−4 k + 0)/( k +1) = −3 ⇒ −4 k = −3 k − 3 ⇒ k = 3 (ratio 3 : 1). [1 Mark] (d) Line BD through (4, −4) and (−3, 0): slope = −4/7 ⇒ 4 x + 7 y + 12 = 0 . [1 Mark]

Question 5

The coordinates of the vertices of $	riangle ABC$ are respectively $(-4, -2)$, $(6, 2)$ and $(4, 6)$. The centroid $G$ of $	riangle ABC$ is:

Accepted Answer

Centroid formula: $G(x, y) = \left(\frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3}\right)$.
$x = \frac{-4 + 6 + 4}{3} = \frac{6}{3} = 2$.
$y = \frac{-2 + 2 + 6}{3} = \frac{6}{3} = 2$.
Coordinates of $G = (2, 2)$. Correct Option: (a) [1 Mark]

Question 6

In what ratio is the line joining $P(5, 3)$ and $Q(-5, 3)$ divided by the y-axis? Also find the coordinates of the point of intersection.

Accepted Answer

Step 1 Let the y-axis divide the segment joining $P(5, 3)$ and $Q(-5, 3)$ in the ratio $k : 1$ at point $R(0, y)$ (since any point on the y-axis has $x = 0$). [1 Mark]
Step 2 By section formula for the x-coordinate:
$$x = \frac{k(-5) + 1(5)}{k + 1} \implies 0 = \frac{-5k + 5}{k + 1} \implies -5k + 5 = 0 \implies 5k = 5 \implies k = 1$$
Thus, the required ratio is $1 : 1$ (the segment is bisected by the y-axis). [1.5 Marks]
Step 3 By section formula for the y-coordinate:
$$y = \frac{1(3) + 1(3)}{1 + 1} = \frac{6}{2} = 3$$
Answer: Ratio = $1 : 1$, Point of intersection = $(0, 3)$. [1.5 Marks]

Question 7

$M$ and $N$ are two points on the X axis and Y axis respectively. $P(3, 2)$ divides the line segment $MN$ in the ratio $2 : 3$. Find: the coordinates of $M$ and $N$ slope of the line $MN$.

Accepted Answer

Step 1 Assign coordinates:
$M$ lies on X-axis $\implies M(x, 0)$.
$N$ lies on Y-axis $\implies N(0, y)$.
Ratio $m_1 : m_2 = 2 : 3$, and $P(3, 2)$ divides $MN$.
Step 2 Apply Section Formula:
$x_P = \frac{m_1 x_2 + m_2 x_1}{m_1 + m_2} \implies 3 = \frac{2(0) + 3(x)}{2 + 3} = \frac{3x}{5} \implies 3x = 15 \implies x = 5$.
$y_P = \frac{m_1 y_2 + m_2 y_1}{m_1 + m_2} \implies 2 = \frac{2(y) + 3(0)}{2 + 3} = \frac{2y}{5} \implies 2y = 10 \implies y = 5$.
$\mathbf{\therefore M(5, 0) \text{ and } N(0, 5)}$. [2 Marks]
Step 3 Calculate slope of line $MN$:
$\text{Slope } m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{5 - 0}{0 - 5} = \frac{5}{-5} = \mathbf{-1}$. [1 Mark]

Question 8

The coordinates of the vertices of $\Delta ABC$ are $A(a, b)$, $B(5, -3)$ and $C(-5, 3)$. The coordinates of the centroid $G$ of $\Delta ABC$ is $G\left(\frac{2}{3}, \frac{1}{3}\right)$. Find the: coordinates of the vertex $A$. equation of median through the vertex $C$.

Accepted Answer

Step 1 (a) Use Centroid Formula $G\left(\frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3}\right)$:
$x_G = \frac{a + 5 + (-5)}{3} = \frac{a}{3} = \frac{2}{3} \implies \mathbf{a = 2}$.
$y_G = \frac{b + (-3) + 3}{3} = \frac{b}{3} = \frac{1}{3} \implies \mathbf{b = 1}$.
$\therefore \mathbf{\text{Coordinates of } A = (2, 1)}$. [2 Marks]
Step 2 (b) Find midpoint $M$ of side $AB$ opposite to vertex $C$:
$A(2, 1)$ and $B(5, -3) \implies M = \left(\frac{2 + 5}{2}, \frac{1 + (-3)}{2}\right) = \left(\frac{7}{2}, -1\right)$. [1 Mark]
Step 3 Equation of median line joining $C(-5, 3)$ and $M\left(\frac{7}{2}, -1\right)$:
$\text{Slope } m = \frac{-1 - 3}{\frac{7}{2} - (-5)} = \frac{-4}{\frac{17}{2}} = -\frac{8}{17}$.
Using point-slope form with $C(-5, 3)$:
$y - 3 = -\frac{8}{17}(x + 5) \implies 17(y - 3) = -8(x + 5) \implies 17y - 51 = -8x - 40$
$\mathbf{8x + 17y - 11 = 0}$. [1 Mark]

Question 9

$A(2, 5)$, $B(-1, 2)$ and $C(5, 8)$ are the vertices of a triangle $ABC$, '$M$' is a point on $AB$ such that $AM : MB = 1 : 2$. Find the co-ordinates of '$M$'. Hence find the equation of the line passing through the points $C$ and $M$.

Accepted Answer

Step 1 Coordinates of $M$ dividing $AB$ in ratio $1 : 2$ ($m_1=1, m_2=2$):
$$x_M = \frac{m_1 x_2 + m_2 x_1}{m_1 + m_2} = \frac{1(-1) + 2(2)}{1 + 2} = \frac{-1 + 4}{3} = \frac{3}{3} = 1$$ $$y_M = \frac{m_1 y_2 + m_2 y_1}{m_1 + m_2} = \frac{1(2) + 2(5)}{1 + 2} = \frac{2 + 10}{3} = \frac{12}{3} = 4$$ $$\mathbf{M = (1, 4)}$$ [2 Marks]
Step 2 Slope of line passing through $C(5, 8)$ and $M(1, 4)$:
$$m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{8 - 4}{5 - 1} = \frac{4}{4} = 1$$ [1 Mark]
Step 3 Equation of line $CM$ using point $M(1, 4)$ and $m = 1$:
$$y - 4 = 1(x - 1) \implies y - 4 = x - 1$$ $$\mathbf{x - y + 3 = 0} \quad (\text{or } y = x + 3)$$ [1 Mark]
Answers: Coordinates of $M = (1, 4)$; Equation of line $CM$: $x - y + 3 = 0$.

Question 10

$P(1, -2)$ is a point on the line segment $A(3, -6)$ and $B(x, y)$ such that $AP : PB$ is equal to $2 : 3$. Find the coordinates of $B$.

Accepted Answer

Step 1 Let $A(3, -6) = (x_1, y_1)$ and $B(x, y) = (x_2, y_2)$ with ratio $m_1 : m_2 = 2 : 3$.
Step 2: Solve for x
$$x_P = \frac{m_1 x_2 + m_2 x_1}{m_1 + m_2} \implies 1 = \frac{2x + 3(3)}{2 + 3}$$ $$1 = \frac{2x + 9}{5} \implies 5 = 2x + 9 \implies 2x = -4 \implies \mathbf{x = -2}$$ [2 Marks]
Step 3: Solve for y
$$y_P = \frac{m_1 y_2 + m_2 y_1}{m_1 + m_2} \implies -2 = \frac{2y + 3(-6)}{2 + 3}$$ $$-2 = \frac{2y - 18}{5} \implies -10 = 2y - 18 \implies 2y = 8 \implies \mathbf{y = 4}$$ [2 Marks]
Answer: The coordinates of $B$ are $(-2, 4)$.

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