Samacheer Kalvi Class 9 Maths Solutions Chapter 5 Coordinate Geometry Exercise 5.3

Question 1. Find the mid-points of the line segment joining the points.
(i) (-2, 3) and (-6, -5)
(ii) (8, -2) and (-8, 0)
(iii) (a, b) and (a + 2b, 2a - b)
(iv) \( \left(\frac{1}{2},\frac{-3}{7}\right) \) and \( \left(\frac{3}{2},\frac{-11}{7}\right) \)
Answer:
(i) To find the midpoint of the line segment joining \( (-2, 3) \) and \( (-6, -5) \), we use the midpoint formula \( \left( \frac{x_1+x_2}{2}, \frac{y_1+y_2}{2} \right) \).
Mid-point \( = \left( \frac{-2+(-6)}{2}, \frac{3+(-5)}{2} \right) \)
\( = \left( \frac{-8}{2}, \frac{-2}{2} \right) \)
\( = (-4,-1) \)
(ii) To find the midpoint of the line segment joining \( (8, -2) \) and \( (-8, 0) \), we apply the same formula.
Mid-point \( = \left( \frac{8+(-8)}{2}, \frac{-2+0}{2} \right) \)
\( = \left( \frac{0}{2}, \frac{-2}{2} \right) \)
\( = (0,-1) \)
(iii) For the points \( (a, b) \) and \( (a + 2b, 2a - b) \), the calculation is as follows:
Mid-point \( = \left( \frac{a+(a+2b)}{2}, \frac{b+(2a-b)}{2} \right) \)
\( = \left( \frac{2a+2b}{2}, \frac{2a}{2} \right) \)
\( = \left( \frac{2(a+b)}{2}, a \right) \)
\( = (a+b, a) \)
(iv) For the fractional coordinates \( \left(\frac{1}{2},\frac{-3}{7}\right) \) and \( \left(\frac{3}{2},\frac{-11}{7}\right) \), we combine the numerators with a common denominator.
Mid-point \( = \left( \frac{\frac{1}{2}+\frac{3}{2}}{2}, \frac{\frac{-3}{7}+\frac{-11}{7}}{2} \right) \)
\( = \left( \frac{\frac{4}{2}}{2}, \frac{\frac{-14}{7}}{2} \right) \)
\( = \left( \frac{2}{2}, \frac{-2}{2} \right) \)
\( = (1,-1) \)
In simple words: The midpoint of a line segment is found by averaging the x-coordinates and averaging the y-coordinates of the two endpoints. This gives you the exact middle point of the line.

🎯 Exam Tip: Remember the midpoint formula \( \left( \frac{x_1+x_2}{2}, \frac{y_1+y_2}{2} \right) \) and be careful with signs when substituting negative coordinates.

Question 2. The centre of a circle is (-4, 2). If one end of the diameter of the circle is (-3, 7) then find the other end.
Answer: Let the other end of the diameter be \( (a, b) \). We know that the center of a circle is the midpoint of its diameter. The given center is \( (-4, 2) \), and one end of the diameter is \( (-3, 7) \).
Using the midpoint formula \( \left( \frac{x_1+x_2}{2}, \frac{y_1+y_2}{2} \right) \), we can set up the equations:
\( (-4, 2) = \left( \frac{-3+a}{2}, \frac{7+b}{2} \right) \)
Equating the x-coordinates:
\( \frac{-3+a}{2} = -4 \)
\( -3+a = -8 \)
\( a = -8+3 \)
\( a = -5 \)
Equating the y-coordinates:
\( \frac{7+b}{2} = 2 \)
\( 7+b = 4 \)
\( b = 4-7 \)
\( b = -3 \)
So, the coordinates of the other end of the diameter are \( (-5, -3) \). The diameter passes through the center, making the center the exact middle of the diameter's endpoints.
In simple words: A circle's center is always exactly halfway between the two ends of any diameter. We use this fact to find the missing endpoint by setting the center as the midpoint of the known end and the unknown end.

🎯 Exam Tip: Understand that the center of a circle acts as the midpoint of any diameter. This property is key to solving problems involving diameters and centers.

Question 3. If the mid-point (x, y) of the line joining (3, 4) and (p, 7) lies on 2x + 2y + 1 = 0, then what will be the value of p?
Answer: First, we find the mid-point \( (x, y) \) of the line segment joining \( (3, 4) \) and \( (p, 7) \).
Using the midpoint formula \( \left( \frac{x_1+x_2}{2}, \frac{y_1+y_2}{2} \right) \):
\( (x, y) = \left( \frac{3+p}{2}, \frac{4+7}{2} \right) \)
\( (x, y) = \left( \frac{3+p}{2}, \frac{11}{2} \right) \)
We are told that this midpoint lies on the line \( 2x + 2y + 1 = 0 \). This means we can substitute the x and y values of the midpoint into the equation of the line.
\( 2\left( \frac{3+p}{2} \right) + 2\left( \frac{11}{2} \right) + 1 = 0 \)
\( 3+p + 11 + 1 = 0 \)
\( p+15 = 0 \)
\( p = -15 \)
The value of p is \( -15 \). This shows how coordinate geometry connects points, lines, and algebraic equations.
In simple words: Find the middle point between the two given points. Then, put the x and y values of this middle point into the line's equation to find the unknown value, p.

🎯 Exam Tip: When a point lies on a line, its coordinates must satisfy the line's equation. This is a fundamental concept in coordinate geometry for finding unknown values.

Question 4. The mid-point of the sides of a triangle are (2, 4), (-2, 3) and (5, 2). Find the coordinates of the vertices of the triangle.
Answer: Let the vertices of the triangle ABC be \( A(x_1, y_1), B(x_2, y_2) \), and \( C(x_3, y_3) \). Let the given midpoints be \( M_1(2, 4), M_2(-2, 3) \), and \( M_3(5, 2) \).
We can assume \( M_1 \) is the midpoint of AB, \( M_2 \) of BC, and \( M_3 \) of AC.

Midpoint of AB: \( (2, 4) = \left( \frac{x_1+x_2}{2}, \frac{y_1+y_2}{2} \right) \)
\( \implies \frac{x_1+x_2}{2} = 2 \implies x_1+x_2 = 4 \) (1)
\( \implies \frac{y_1+y_2}{2} = 4 \implies y_1+y_2 = 8 \) (2)

Midpoint of BC: \( (-2, 3) = \left( \frac{x_2+x_3}{2}, \frac{y_2+y_3}{2} \right) \)
\( \implies \frac{x_2+x_3}{2} = -2 \implies x_2+x_3 = -4 \) (3)
\( \implies \frac{y_2+y_3}{2} = 3 \implies y_2+y_3 = 6 \) (4)

Midpoint of AC: \( (5, 2) = \left( \frac{x_1+x_3}{2}, \frac{y_1+y_3}{2} \right) \)
\( \implies \frac{x_1+x_3}{2} = 5 \implies x_1+x_3 = 10 \) (5)
\( \implies \frac{y_1+y_3}{2} = 2 \implies y_1+y_3 = 4 \) (6)

To find the x-coordinates of the vertices, we add equations (1), (3), and (5):
\( (x_1+x_2) + (x_2+x_3) + (x_1+x_3) = 4 + (-4) + 10 \)
\( 2x_1 + 2x_2 + 2x_3 = 10 \)
\( 2(x_1+x_2+x_3) = 10 \implies x_1+x_2+x_3 = 5 \) (7)
Substitute \( x_1+x_2 = 4 \) (from eq 1) into eq (7):
\( 4 + x_3 = 5 \implies x_3 = 1 \)
Substitute \( x_2+x_3 = -4 \) (from eq 3) into eq (7):
\( x_1 + (-4) = 5 \implies x_1 = 9 \)
Substitute \( x_1+x_3 = 10 \) (from eq 5) into eq (7):
\( x_2 + 10 = 5 \implies x_2 = -5 \)
So, the x-coordinates are \( x_1 = 9, x_2 = -5, x_3 = 1 \).

To find the y-coordinates of the vertices, we add equations (2), (4), and (6):
\( (y_1+y_2) + (y_2+y_3) + (y_1+y_3) = 8 + 6 + 4 \)
\( 2y_1 + 2y_2 + 2y_3 = 18 \)
\( 2(y_1+y_2+y_3) = 18 \implies y_1+y_2+y_3 = 9 \) (8)
Substitute \( y_1+y_2 = 8 \) (from eq 2) into eq (8):
\( 8 + y_3 = 9 \implies y_3 = 1 \)
Substitute \( y_2+y_3 = 6 \) (from eq 4) into eq (8):
\( y_1 + 6 = 9 \implies y_1 = 3 \)
Substitute \( y_1+y_3 = 4 \) (from eq 6) into eq (8):
\( y_2 + 4 = 9 \implies y_2 = 5 \)
So, the y-coordinates are \( y_1 = 3, y_2 = 5, y_3 = 1 \).

The vertices of the triangle are \( A(9, 3), B(-5, 5) \), and \( C(1, 1) \). This method of using a system of equations is very common in coordinate geometry.
In simple words: We know the middle points of each side of the triangle. We set up equations using the midpoint formula for each side. By adding these equations, we can find the sum of all x-coordinates and all y-coordinates. Then we use substitution to find each individual x and y coordinate of the triangle's corners.

🎯 Exam Tip: This problem requires solving a system of linear equations. Clearly label your equations and substitute carefully to avoid errors, especially with signs.

Question 5. O (0, 0) is the centre of a circle whose one chord is AB, where the points A and B (8, 6) and (10, 0) respectively. OD is the perpendicular from the centre to the chord AB. Find the coordinates of the mid-point of OD.
Answer: We are given that O is the center \( (0, 0) \). A chord AB has endpoints \( A(8, 6) \) and \( B(10, 0) \). OD is perpendicular to the chord AB. A key property of circles is that a perpendicular drawn from the center to a chord bisects the chord. This means D is the midpoint of AB.

First, find the coordinates of D (midpoint of AB):
Using the midpoint formula \( \left( \frac{x_1+x_2}{2}, \frac{y_1+y_2}{2} \right) \):
\( D = \left( \frac{8+10}{2}, \frac{6+0}{2} \right) \)
\( D = \left( \frac{18}{2}, \frac{6}{2} \right) \)
\( D = (9, 3) \)

Next, we need to find the midpoint of the line segment OD. O is \( (0, 0) \) and D is \( (9, 3) \).
Using the midpoint formula again:
Midpoint of OD \( = \left( \frac{0+9}{2}, \frac{0+3}{2} \right) \)
Midpoint of OD \( = \left( \frac{9}{2}, \frac{3}{2} \right) \)
So, the coordinates of the midpoint of OD are \( \left( \frac{9}{2}, \frac{3}{2} \right) \). This problem combines geometric properties with coordinate calculations.
In simple words: First, find the middle point of chord AB. This is point D. Then, find the middle point between the center O (0,0) and point D. That is the final answer.

🎯 Exam Tip: Always remember the geometric property that a perpendicular from the center to a chord bisects the chord. This allows you to find D using the midpoint formula.

Question 6. The points A(-5, 4), B(-1, -2) and C(5, 2) are the vertices of an isosceles right-angled triangle where the right angle is at B. Find the coordinates of D so that ABCD is a square.
Answer: In a square ABCD, the diagonals bisect each other, meaning they share the same midpoint. We are given vertices \( A(-5, 4), B(-1, -2) \), and \( C(5, 2) \). Let the coordinates of the fourth vertex D be \( (a, b) \).

First, find the midpoint of diagonal AC:
Midpoint of AC \( = \left( \frac{x_A+x_C}{2}, \frac{y_A+y_C}{2} \right) \)
\( = \left( \frac{-5+5}{2}, \frac{4+2}{2} \right) \)
\( = \left( \frac{0}{2}, \frac{6}{2} \right) \)
\( = (0, 3) \)

Next, find the midpoint of diagonal BD:
Midpoint of BD \( = \left( \frac{x_B+x_D}{2}, \frac{y_B+y_D}{2} \right) \)
\( = \left( \frac{-1+a}{2}, \frac{-2+b}{2} \right) \)

Since the midpoints of the diagonals are the same for a square, we equate the coordinates:
\( \left( \frac{-1+a}{2}, \frac{-2+b}{2} \right) = (0, 3) \)
Equating the x-coordinates:
\( \frac{-1+a}{2} = 0 \)
\( -1+a = 0 \)
\( a = 1 \)
Equating the y-coordinates:
\( \frac{-2+b}{2} = 3 \)
\( -2+b = 6 \)
\( b = 6+2 \)
\( b = 8 \)
Therefore, the coordinates of vertex D are \( (1, 8) \). This problem shows how properties of quadrilaterals translate into coordinate relationships.
In simple words: For a square, the center point of both diagonals is the same. We find the middle of the known diagonal (AC). Then we use this midpoint and the known point B to find the missing point D.

🎯 Exam Tip: The property that diagonals of a square bisect each other is crucial. If you forget it, you might try distance formulas, which would be much longer.

Question 7. The points A (-3, 6), B (0, 7) and C (1, 9) are the mid points of the sides DE, EF and FD of a triangle DEF. Show that the quadrilateral ABCD is a parallelogram.
Answer: Let the vertices of triangle DEF be \( D(x_1, y_1), E(x_2, y_2) \), and \( F(x_3, y_3) \). We are given that A, B, C are the midpoints of the sides DE, EF, FD respectively.

For midpoint A \( (-3, 6) \) of DE:
\( (-3, 6) = \left( \frac{x_1+x_2}{2}, \frac{y_1+y_2}{2} \right) \)
\( x_1+x_2 = -6 \) (1)
\( y_1+y_2 = 12 \) (2)

For midpoint B \( (0, 7) \) of EF:
\( (0, 7) = \left( \frac{x_2+x_3}{2}, \frac{y_2+y_3}{2} \right) \)
\( x_2+x_3 = 0 \) (3)
\( y_2+y_3 = 14 \) (4)

For midpoint C \( (1, 9) \) of FD:
\( (1, 9) = \left( \frac{x_1+x_3}{2}, \frac{y_1+y_3}{2} \right) \)
\( x_1+x_3 = 2 \) (5)
\( y_1+y_3 = 18 \) (6)

Now, we find the coordinates of D. Add equations (1), (3), and (5):
\( (x_1+x_2) + (x_2+x_3) + (x_1+x_3) = -6 + 0 + 2 \)
\( 2x_1 + 2x_2 + 2x_3 = -4 \)
\( x_1+x_2+x_3 = -2 \) (7)
Substitute \( x_1+x_2 = -6 \) (from eq 1) into eq (7):
\( -6 + x_3 = -2 \implies x_3 = 4 \)
Substitute \( x_2+x_3 = 0 \) (from eq 3) into eq (7):
\( x_1 + 0 = -2 \implies x_1 = -2 \)
Substitute \( x_1+x_3 = 2 \) (from eq 5) into eq (7):
\( x_2 + 2 = -2 \implies x_2 = -4 \)
So, \( D(x_1, y_1) = (-2, y_1) \).

Now for y-coordinates. Add equations (2), (4), and (6):
\( (y_1+y_2) + (y_2+y_3) + (y_1+y_3) = 12 + 14 + 18 \)
\( 2y_1 + 2y_2 + 2y_3 = 44 \)
\( y_1+y_2+y_3 = 22 \) (8)
Substitute \( y_1+y_2 = 12 \) (from eq 2) into eq (8):
\( 12 + y_3 = 22 \implies y_3 = 10 \)
Substitute \( y_2+y_3 = 14 \) (from eq 4) into eq (8):
\( y_1 + 14 = 22 \implies y_1 = 8 \)
Substitute \( y_1+y_3 = 18 \) (from eq 6) into eq (8):
\( y_2 + 18 = 22 \implies y_2 = 4 \)
So, the vertex D is \( (-2, 8) \).

Now we need to show that quadrilateral ABCD is a parallelogram. We can do this by showing that its diagonals AC and BD have the same midpoint.
Given points for ABCD are: \( A(-3, 6), B(0, 7), C(1, 9) \), and \( D(-2, 8) \).

Midpoint of diagonal AC:
\( = \left( \frac{-3+1}{2}, \frac{6+9}{2} \right) = \left( \frac{-2}{2}, \frac{15}{2} \right) = \left( -1, \frac{15}{2} \right) \)

Midpoint of diagonal BD:
\( = \left( \frac{0+(-2)}{2}, \frac{7+8}{2} \right) = \left( \frac{-2}{2}, \frac{15}{2} \right) = \left( -1, \frac{15}{2} \right) \)

Since the midpoint of diagonal AC is equal to the midpoint of diagonal BD, the quadrilateral ABCD is a parallelogram. This demonstrates how finding missing coordinates helps prove geometric properties.
In simple words: First, we use the midpoint formula to find the coordinates of D, E, and F based on the given midpoints A, B, C. Once we have D, we then check if the diagonals of ABCD cut each other in the middle. If they do, then ABCD is a parallelogram.

🎯 Exam Tip: To prove a quadrilateral is a parallelogram using coordinates, the easiest method is to show that the midpoints of its two diagonals coincide. Alternatively, you could show that opposite sides are parallel by comparing slopes.

Question 8. A(-3, 2), B(3, 2) and C(-3, -2) are the vertices of the right triangle, right angled at A. Show that the mid point of the hypotenuse is equidistant from the vertices.
Answer: Given the vertices of the right triangle are \( A(-3, 2), B(3, 2) \), and \( C(-3, -2) \). Since the right angle is at A, the hypotenuse is BC. Let D be the midpoint of the hypotenuse BC.

First, find the coordinates of D, the midpoint of BC:
Using the midpoint formula \( \left( \frac{x_1+x_2}{2}, \frac{y_1+y_2}{2} \right) \):
\( D = \left( \frac{3+(-3)}{2}, \frac{2+(-2)}{2} \right) \)
\( D = \left( \frac{0}{2}, \frac{0}{2} \right) \)
\( D = (0,0) \)

Next, we calculate the distance from D to each of the vertices A, B, and C using the distance formula \( d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2} \).

Distance DA (from \( D(0,0) \) to \( A(-3,2) \)):
\( DA = \sqrt{(-3-0)^2 + (2-0)^2} \)
\( DA = \sqrt{(-3)^2 + (2)^2} \)
\( DA = \sqrt{9+4} \)
\( DA = \sqrt{13} \)

Distance DB (from \( D(0,0) \) to \( B(3,2) \)):
\( DB = \sqrt{(3-0)^2 + (2-0)^2} \)
\( DB = \sqrt{(3)^2 + (2)^2} \)
\( DB = \sqrt{9+4} \)
\( DB = \sqrt{13} \)

Distance DC (from \( D(0,0) \) to \( C(-3,-2) \)):
\( DC = \sqrt{(-3-0)^2 + (-2-0)^2} \)
\( DC = \sqrt{(-3)^2 + (-2)^2} \)
\( DC = \sqrt{9+4} \)
\( DC = \sqrt{13} \)

Since \( DA = DB = DC = \sqrt{13} \), the midpoint of the hypotenuse is equidistant from all three vertices of the right triangle. This is a special property of right-angled triangles, where the circumcenter is the midpoint of the hypotenuse.
In simple words: First, find the middle point of the longest side of the right triangle (the hypotenuse). Then, measure the distance from this middle point to all three corners of the triangle. If all three distances are the same, it means the middle point of the hypotenuse is equally far from every corner.

🎯 Exam Tip: This problem illustrates a key geometric theorem: the midpoint of the hypotenuse of a right-angled triangle is the circumcenter, meaning it is equidistant from all three vertices. Knowing this property can help you quickly verify your calculations.