Extra Practice Questions
Coordinate Geometry (R.D. Sharma Level)
1. Determine the ratio in which the y-axis divides the line segment joining the points (5, –6) and (–1, –4). Also find the point of intersection.
Let the ratio be k:1. Any point on the y-axis has coordinates (0, y).
Using the section formula for the x-coordinate:
0 = k(-1) + 1(5)k + 1 => 0 = -k + 5 => k = 5.
The ratio is 5:1.
Now find the y-coordinate:
y = 5(-4) + 1(-6)5 + 1 = -20 – 66 = -26/6 = -13/3.
Answer: The ratio is 5:1 and the point is (0, -13/3).
2. Find the coordinates of the centroid of a triangle whose vertices are (3, –5), (–7, 4), and (10, –2).
The centroid (G) of a triangle with vertices (x₁, y₁), (x₂, y₂), (x₃, y₃) is given by:
G(x, y) = (
x₁ + x₂ + x₃3,
y₁ + y₂ + y₃3
)
x = 3 + (-7) + 103 = 63 = 2.
y = -5 + 4 + (-2)3 = -33 = -1.
Answer: The centroid is (2, -1).
… (18 more questions follow)
3. If the points A(6, 1), B(8, 2), C(9, 4) and D(p, 3) are the vertices of a parallelogram, taken in order, find the value of p.
The diagonals of a parallelogram bisect each other. So, Midpoint of AC = Midpoint of BD.
Midpoint of AC = (6+92, 1+42) = (15/2, 5/2).
Midpoint of BD = (8+p2, 2+32) = (8+p2, 5/2).
Equating the x-coordinates: 15/2 = (8+p)/2 => 15 = 8+p => p=7.
Answer: p = 7.
4. Find the value of k if the points A(2, 3), B(4, k) and C(6, –3) are collinear.
For collinear points, the area of the triangle formed by them is 0.
Area = 1/2 [x₁(y₂-y₃) + x₂(y₃-y₁) + x₃(y₁-y₂)] = 0.
2(k – (-3)) + 4(-3 – 3) + 6(3 – k) = 0
2(k+3) + 4(-6) + 6(3-k) = 0
2k + 6 – 24 + 18 – 6k = 0
-4k = 0 => k = 0.
Answer: k = 0.
5. Find the coordinates of the point on the line y = x which is equidistant from the points (3, 4) and (1, -2).
Let the point be P(a, a) since it lies on the line y=x. Let A=(3,4), B=(1,-2).
Given PA² = PB².
(a – 3)² + (a – 4)² = (a – 1)² + (a – (-2))²
(a² – 6a + 9) + (a² – 8a + 16) = (a² – 2a + 1) + (a² + 4a + 4)
2a² – 14a + 25 = 2a² + 2a + 5
-14a + 25 = 2a + 5 => 20 = 16a => a = 20/16 = 5/4.
Answer: The point is (5/4, 5/4).
6. Find the area of the quadrilateral whose vertices, taken in order, are (–4, –2), (–3, –5), (3, –2) and (2, 3).
Let A(-4,-2), B(-3,-5), C(3,-2), D(2,3). We find the area of ΔABC and ΔACD.
Area(ΔABC) = 1/2 | -4(-5 – (-2)) + (-3)(-2 – (-2)) + 3(-2 – (-5)) |
= 1/2 | -4(-3) – 3(0) + 3(3) | = 1/2 |12 + 9| = 21/2 sq. units.
Area(ΔACD) = 1/2 | -4(-2 – 3) + 3(3 – (-2)) + 2(-2 – (-2)) |
= 1/2 | -4(-5) + 3(5) + 2(0) | = 1/2 |20 + 15| = 35/2 sq. units.
Area of quadrilateral = Area(ΔABC) + Area(ΔACD) = 21/2 + 35/2 = 56/2 = 28.
Answer: The area is 28 square units.
7. If two vertices of a parallelogram are (3, 2), (–1, 0) and the diagonals intersect at (2, –5), find the other two vertices.
Let A=(3,2), C=(x,y). The midpoint of AC is the intersection point O(2,-5).
(3+x)/2 = 2 => 3+x=4 => x=1.
(2+y)/2 = -5 => 2+y=-10 => y=-12. So, vertex C is (1, -12).
Let B=(-1,0), D=(p,q). The midpoint of BD is also O(2,-5).
(-1+p)/2 = 2 => -1+p=4 => p=5.
(0+q)/2 = -5 => q=-10. So, vertex D is (5, -10).
Answer: The other two vertices are (1, -12) and (5, -10).
8. The line segment joining the points (3, -4) and (1, 2) is trisected at the points P and Q. If the coordinates of P and Q are (p, -2) and (5/3, q) respectively, find the values of p and q.
Let A=(3,-4) and B=(1,2). P divides AB in ratio 1:2. Q divides in ratio 2:1.
Coordinates of P: x = (1(1)+2(3))/(1+2) = 7/3. y = (1(2)+2(-4))/3 = -6/3 = -2.
So P = (7/3, -2). Given P=(p,-2), this means p = 7/3.
Coordinates of Q: x = (2(1)+1(3))/(2+1) = 5/3. y = (2(2)+1(-4))/3 = 0/3 = 0.
So Q = (5/3, 0). Given Q=(5/3,q), this means q = 0.
Answer: p = 7/3, q = 0.
9. Find the coordinates of the circumcentre of the triangle whose vertices are (8, 6), (8, -2) and (2, -2).
Let the vertices be A(8,6), B(8,-2), C(2,-2). Let circumcentre be P(x,y). PA=PB=PC.
PA² = PB² => (x-8)²+(y-6)² = (x-8)²+(y+2)² => (y-6)²=(y+2)² => y²-12y+36=y²+4y+4 => 16y=32 => y=2.
PB² = PC² => (x-8)²+(y+2)² = (x-2)²+(y+2)² => (x-8)²=(x-2)² => x²-16x+64=x²-4x+4 => 12x=60 => x=5.
Answer: The circumcentre is (5, 2).
10. Find the ratio in which the line 2x + 3y – 5 = 0 divides the line segment which joins the points (8, -9) and (2, 1). Also find the coordinates of the point of division.
Let the ratio be k:1. The point of division P(x,y) is:
x = (2k+8)/(k+1), y = (k-9)/(k+1).
This point lies on the line 2x+3y-5=0. Substitute x and y:
2[(2k+8)/(k+1)] + 3[(k-9)/(k+1)] – 5 = 0.
Multiply by (k+1): 2(2k+8) + 3(k-9) – 5(k+1) = 0.
4k+16 + 3k-27 – 5k-5 = 0 => 2k – 16 = 0 => k=8.
Ratio is 8:1. Point of division:
x = (2(8)+8)/9 = 24/9 = 8/3.
y = (8-9)/9 = -1/9.
Answer: Ratio is 8:1, point is (8/3, -1/9).
11. If the mid-points of the sides of a triangle are (1, 2), (0, -1) and (2, -1), find the coordinates of its vertices.
Let vertices be A(x₁,y₁), B(x₂,y₂), C(x₃,y₃). Let midpoints be D(1,2) on BC, E(0,-1) on AC, F(2,-1) on AB.
x₂+x₃=2, x₁+x₃=0, x₁+x₂=4. Adding these: 2(x₁+x₂+x₃)=6 => x₁+x₂+x₃=3.
x₁=3-2=1, x₂=3-0=3, x₃=3-4=-1.
y₂+y₃=4, y₁+y₃=-2, y₁+y₂=-2. Adding these: 2(y₁+y₂+y₃)=0 => y₁+y₂+y₃=0.
y₁=0-4=-4, y₂=0-(-2)=2, y₃=0-(-2)=2.
Answer: The vertices are (1, -4), (3, 2), and (-1, 2).
12. Prove that the points (a, a), (-a, -a) and (-√3a, √3a) are the vertices of an equilateral triangle.
Let A=(a,a), B=(-a,-a), C=(-√3a, √3a).
AB² = (-2a)²+(-2a)² = 4a²+4a²=8a².
BC² = (-√3a+a)²+(√3a+a)² = a²(-\√3+1)²+a²(√3+1)² = a²(3-2√3+1 + 3+2√3+1) = a²(8) = 8a².
AC² = (-√3a-a)²+(√3a-a)² = a²(-√3-1)²+a²(√3-1)² = a²(3+2√3+1 + 3-2√3+1) = a²(8) = 8a².
Since AB²=BC²=AC²=8a², all sides are equal. Hence, it’s an equilateral triangle. (Proved)
13. Point P divides the line segment joining R(–1, 3) and S(9, 8) in the ratio k:1. If P lies on the line x – y + 2 = 0, find the value of k.
Coordinates of P are: x = (9k-1)/(k+1), y = (8k+3)/(k+1).
P lies on x-y+2=0. Substitute the coordinates:
[(9k-1)/(k+1)] – [(8k+3)/(k+1)] + 2 = 0.
Multiply by (k+1): (9k-1) – (8k+3) + 2(k+1) = 0.
9k-1-8k-3+2k+2=0 => 3k-2=0 => k=2/3.
Answer: k = 2/3.
14. Find the area of the triangle formed by joining the mid-points of the sides of the triangle whose vertices are (0, -1), (2, 1) and (0, 3). Find the ratio of this area to the area of the given triangle.
Let A(0,-1), B(2,1), C(0,3). Midpoints are D(1,2), E(0,1), F(1,0).
Area(ΔDEF) = 1/2 |1(1-0)+0(0-2)+1(2-1)| = 1/2 |1+0+1| = 1 sq. unit.
Area(ΔABC) = 1/2 |0(1-3)+2(3-(-1))+0(-1-1)| = 1/2 |0+8+0| = 4 sq. units.
Ratio = Area(ΔDEF)/Area(ΔABC) = 1/4.
Answer: Area of smaller triangle is 1 sq. unit; the ratio is 1:4.
15. If A(–5, 7), B(–4, –5), C(–1, –6) and D(4, 5) are the vertices of a quadrilateral, find the area of the quadrilateral ABCD.
We find the area of ΔABC and ΔACD.
Area(ΔABC) = 1/2 | -5(-5 – (-6)) + (-4)(-6 – 7) + (-1)(7 – (-5)) |
= 1/2 | -5(1) – 4(-13) – 1(12) | = 1/2 |-5+52-12| = 1/2(35) = 17.5 sq. units.
Area(ΔACD) = 1/2 | -5(-6 – 5) + (-1)(5 – 7) + 4(7 – (-6)) |
= 1/2 | -5(-11) – 1(-2) + 4(13) | = 1/2 |55+2+52| = 1/2(109) = 54.5 sq. units.
Area of quadrilateral = 17.5 + 54.5 = 72.
Answer: The area is 72 square units.
16. The three vertices of a parallelogram taken in order are (–1, 0), (3, 1) and (2, 2). Find the coordinates of the fourth vertex.
Let A(-1,0), B(3,1), C(2,2), D(x,y).
Midpoint of AC = ((-1+2)/2, (0+2)/2) = (1/2, 1).
Midpoint of BD = ((3+x)/2, (1+y)/2).
Equating coordinates: (3+x)/2=1/2 => 3+x=1 => x=-2.
(1+y)/2=1 => 1+y=2 => y=1.
Answer: The fourth vertex is (-2, 1).
17. Find the value of ‘p’ for which the points (–5, 1), (1, p) and (4, –2) are collinear.
If points are collinear, slope of AB = slope of BC.
Slope of AB = (p-1)/(1-(-5)) = (p-1)/6.
Slope of BC = (-2-p)/(4-1) = (-2-p)/3.
(p-1)/6 = (-2-p)/3 => p-1 = 2(-2-p) => p-1 = -4-2p => 3p = -3 => p = -1.
Answer: p = -1.
18. If the point P(x, y) is equidistant from the points A(5, 1) and B(–1, 5), prove that 3x = 2y.
Given PA = PB, so PA² = PB².
(x-5)² + (y-1)² = (x+1)² + (y-5)²
x²-10x+25 + y²-2y+1 = x²+2x+1 + y²-10y+25
-10x – 2y + 26 = 2x – 10y + 26
-10x – 2y = 2x – 10y => 8y = 12x => 2y = 3x.
(Proved)
19. If (a, b) is the mid-point of the line segment joining the points A(10, -6) and B(k, 4) and a – 2b = 18, find the value of k and the distance AB.
Midpoint (a,b) = ((10+k)/2, (-6+4)/2) = ((10+k)/2, -1).
So, b = -1.
Given a – 2b = 18. Substitute b=-1: a – 2(-1) = 18 => a+2=18 => a=16.
Now find k: a = (10+k)/2 => 16 = (10+k)/2 => 32=10+k => k=22.
Points are A(10,-6) and B(22,4).
Distance AB = √[(22-10)²+(4-(-6))²] = √[12²+10²] = √144+100 = √244 = 2√61.
Answer: k=22, Distance AB = 2√61.
20. Two vertices of a triangle are (–1, 4) and (5, 2). If its centroid is (0, –3), find the third vertex.
Let vertices be A(-1,4), B(5,2), C(x,y). Centroid G(0,-3).
x-coordinate of G: 0 = (-1+5+x)/3 => 0 = 4+x => x=-4.
y-coordinate of G: -3 = (4+2+y)/3 => -9 = 6+y => y=-15.
Answer: The third vertex is (-4, -15).
