$$\mathrm{Co}-\mathrm{Ordinate Geometry}$$
Distance formula
2007,2008,2009,2009,
2010,2010,2011,2011,2011,
2011, 2012, 2012, 2012, 2013,
2013,2014,2015,2015,2016,
2016,2017,2018,2018,2018,
2018,2020,2021,2022, 2023, 2024, 2025
Section formula
2007,2008,2009,2009,
2010,2010,2010,2010,2011,
2011, 2012, 2012, 2012, 2012,
2013,2014,2015,2015,2016,
2016,2016,2016,2017,2018,
2019, 2020,2021,2022, 2023, 2024, 2025
Geometry
mid-point formula
2009,2010,2011,
2012,2017
1. Show that the points $$(7,10),(-2,5)$$ and $$(3,-4)$$ are the vertices of an isosceles right triangle. [CBSE 2007-3M]
Solution:
Step 1: Find distances between all points
Let A = (7, 10), B = (-2, 5), C = (3, -4)
Distance AB = √[(7 – (-2))² + (10 – 5)²]
= √[(9)² + (5)²] = √[81 + 25]
= √106
Distance BC = √[(-2 – 3)² + (5 – (-4))²]
= √[(-5)² + (9)²] = √[25 + 81]
= √106
Distance AC = √[(7 – 3)² + (10 – (-4))²]
= √[(4)² + (14)²] = √[16 + 196]
= √212
Step 2: Check if triangle is isosceles
Since AB = BC = √106, two sides are equal.
Therefore, triangle ABC is isosceles.
Step 3: Check if triangle is right-angled
Using Pythagoras theorem:
AB² + BC² = (√106)² + (√106)² = 106 + 106 = 212
AC² = (√212)² = 212
Since AB² + BC² = AC², triangle ABC is right-angled at B.
Conclusion: The points form an isosceles right triangle.
Answer: The points (7, 10), (-2, 5) and (3, -4) form an isosceles right triangle with equal sides AB and BC, and right angle at B.
2. In what ratio does the line $$x-y-2=0$$ divides the line segment joining $$(3,-1)$$ and ( 8,9 )? [CBSE 2007-3M]
Solution:
Step 1: Let the ratio be \(k:1\)
Let the line \(x – y – 2 = 0\) divide the line segment joining \(A(3, -1)\) and \(B(8, 9)\) in the ratio \(k:1\)
Step 2: Find coordinates of division point
Using section formula:
\(x\)-coordinate = \(\frac{k \times 8 + 1 \times 3}{k + 1} = \frac{8k + 3}{k + 1}\)
\(y\)-coordinate = \(\frac{k \times 9 + 1 \times (-1)}{k + 1} = \frac{9k – 1}{k + 1}\)
Step 3: Point lies on the line \(x – y – 2 = 0\)
Substitute coordinates into the line equation:
\(\frac{8k + 3}{k + 1} – \frac{9k – 1}{k + 1} – 2 = 0\)
Step 4: Simplify the equation
Multiply both sides by \((k + 1)\):
\((8k + 3) – (9k – 1) – 2(k + 1) = 0\)
\(8k + 3 – 9k + 1 – 2k – 2 = 0\)
\(-3k + 2 = 0\)
\(-3k = -2\)
\(k = \frac{2}{3}\)
Step 5: Find the ratio
Since \(k = \frac{2}{3}\), the ratio is \(\frac{2}{3} : 1 = 2 : 3\)
Answer: The line \(x – y – 2 = 0\) divides the line segment joining \((3, -1)\) and \((8, 9)\) in the ratio \(2:3\)
3. For what value of $$p$$, are points $$(2,1),(p,-1)$$ and $$(-1,3)$$ collinear? [CBSE 2008-2M]
Solution:
Step 1: Condition for collinearity
Three points are collinear if the area of triangle formed by them is zero.
Area of triangle = \(\frac{1}{2}|x_1(y_2 – y_3) + x_2(y_3 – y_1) + x_3(y_1 – y_2)| = 0\)
Step 2: Apply the formula
Let \(A(2, 1)\), \(B(p, -1)\), \(C(-1, 3)\)
Area = \(\frac{1}{2}|2(-1 – 3) + p(3 – 1) + (-1)(1 – (-1))| = 0\)
Step 3: Simplify the expression
\(\frac{1}{2}|2(-4) + p(2) + (-1)(2)| = 0\)
\(\frac{1}{2}|-8 + 2p – 2| = 0\)
\(\frac{1}{2}|2p – 10| = 0\)
Step 4: Solve for \(p\)
\(|2p – 10| = 0\)
\(2p – 10 = 0\)
\(2p = 10\)
\(p = 5\)
Answer: The points \((2, 1)\), \((p, -1)\) and \((-1, 3)\) are collinear when \(p = 5\)
4. If the distances of $$\mathrm{P}(\mathrm{x}, \mathrm{y})$$ from the points $$\mathrm{A}(3,6)$$ and $$\mathrm{B}(-3,4)$$ are equal, prove that 3 x $$+\mathrm{y}=5$$. [CBSE 2008-3M]
Solution:
Step 1: Write the distance condition
Given: \(PA = PB\)
Distance formula: \(PA = \sqrt{(x – 3)^2 + (y – 6)^2}\)
Distance formula: \(PB = \sqrt{(x + 3)^2 + (y – 4)^2}\)
Step 2: Square both sides to eliminate square roots
Since \(PA = PB\), then:
\((x – 3)^2 + (y – 6)^2 = (x + 3)^2 + (y – 4)^2\)
Step 3: Expand both sides
Left side: \((x – 3)^2 + (y – 6)^2\)
\(= x^2 – 6x + 9 + y^2 – 12y + 36\)
Right side: \((x + 3)^2 + (y – 4)^2\)
\(= x^2 + 6x + 9 + y^2 – 8y + 16\)
Step 4: Simplify the equation
\(x^2 – 6x + 9 + y^2 – 12y + 36\)
\(= x^2 + 6x + 9 + y^2 – 8y + 16\)
Cancel \(x^2\), \(y^2\), and 9 from both sides:
\(-6x – 12y + 36 = 6x – 8y + 16\)
Step 5: Rearrange terms
\(-6x – 12y + 36 – 6x + 8y – 16 = 0\)
\(-12x – 4y + 20 = 0\)
Step 6: Simplify the equation
Divide all terms by \(-4\):
\(3x + y – 5 = 0\)
Therefore, \(3x + y = 5\)
Answer: Proved that if distances of \(P(x, y)\) from \(A(3, 6)\) and \(B(-3, 4)\) are equal, then \(3x + y = 5\)
5. Determine the ratio in which the line $$3 x + 4 y-9=0$$ divides the line-segment joining the points $$(1,3)$$ and $$(2,7)$$. [CBSE 2008-3M]
6. Find the value of $$\boldsymbol{a}$$ so that the point $$(3, \boldsymbol{a})$$ lies on the line represented by $$2 x-3 y=5$$. [CBSE 2009-1M]
7. Find the point on $$y$$-axis which is equidistant from the points $$(5,-2)$$ and (-3, 2). [CBSE 2009-3M]
8. The line segment joining the points $$A(2,1)$$ and $$B(5,-8)$$ is trisected at the points $$P$$ and Q such that P is nearer to A . If P also lies on the line given by $$2 x-y+k=0$$, find the value of $$k$$. [CBSE 2009-3M]
9. If $$\mathrm{P}(x, y)$$ is any point on the line joining the points $$\mathrm{A}(a, 0)$$ and $$\mathrm{B}(0, b)$$, then show that $$\frac{x}{a}+\frac{y}{b}=1$$. [CBSE 2009-3M]
10. If the points $$\mathrm{A}(4,3)$$ and $$\mathrm{B}(\mathrm{x}, 5)$$ are on the circle with the centre $$\mathrm{O}(2,3)$$, find the value of x . [CBSE 2009-2M]
11. Find the ratio in which the point $$(2, y)$$ divides the line segment joining the points $$A(-2,2)$$ and $$B(3,7)$$. Also find the value of $$y$$. [CBSE 2009-3M]
12. Find the area of the quadrilateral $$A B C D$$ whose vertices are $$A(-4,-2), B(-3,-5)$$, C $$(3,-2)$$ and D $$(2,3)$$. [CBSE 2009-3M]
13. If $$P(2, p)$$ is the mid-point of the line segment joining the points $$A(6,-5)$$ and $$B(-2,11)$$, find the value of p. CBSE2010-1M
14. If $$A(1,2), B(4,3)$$ and $$C(6,6)$$ are the vertices of parallelogram $$A B C D$$, find the coordinates of vertex D. CBSE2010-1M
15. Point $$P$$ divides the line segment joining the points $$A(2,1)$$ and $$B(5,-8)$$ such that $$\frac{A P}{A B}=\frac{1}{3}$$. If p lies on the line $$2 \mathrm{x}-\mathrm{y}+\mathrm{k}=0$$, find the value of k . CBSE2010-3M
16. If $$R(x, y)$$ is a point on the line segment joining the points $$P(a, b)$$ and $$Q(b, a)$$, then prove that $$\mathrm{x}+\mathrm{y}=\mathrm{a}+\mathrm{b}$$. CBSE2010-3M
17. Find the distance between the points, $$A(2 a, 6 a)$$ and $$B(2 a+\sqrt{3} a, 5 a)$$.CBSE2010-1M
18. Find the value of $$k$$ if $$P(4,-2)$$ is the midpoint of line segment joining the points $$A(5 k, 3)$$ and $$B(-k,-7)$$. CBSE2010-1M
19. If point $$P\left(\frac{1}{2}, y\right)$$ lies on line segment joining the points $$A(3,-5)$$ and $$B(-7,9)$$, then find the ratio in which P divides AB . Also find the value of y . CBSE2010-3M
20. Find the value of $$k$$ for which the points $$A(9, k), B(4,-2)$$ and $$C(3,-3)$$ are collinear. CBSE2010-3M
21. Find the quadrant in which the point $$P$$ that divides the line segment joining the points $$A(2,-5)$$ and $$B(5,2)$$ in the ratio $$2: 3$$. CBSE2011-1M
22. The mid-points of line segment $$A B$$ is the point $$P(0,4)$$. If the coordinates of $$B$$ are $$(2,-3)$$ then find the coordinates of A. CBSE2011-1M
23. Find the values of $$x$$ for which the distance between the points $$P(x, 4)$$ and $$Q(9,10)$$ is 10 units. CBSE2011-2M
24. If $$(3,3),(6, y),(x, 7)$$ and $$(5,6)$$ are the vertices of parallelogram taken in order then find