The distance of the origin from the centroid of the triangle whose two sides have the equations
x – 2y + 1 = 0 and 2x – y – 1 = 0 and whose orthocenter is (7/3, 7/3) is :
x – 2y + 1 = 0 and 2x – y – 1 = 0 and whose orthocenter is (7/3, 7/3) is :
- √2
- 2
- 2√2
- 4
The Correct Option is A
Solution and Explanation
The correct answer is (C) : \(2\sqrt2\)
AB = x – 2y + 1 = 0
AC = 2x – y - 1 = 0
So A(1, 1)
Altitude from B is BH \(= x + 2y – 7 = 0 ⇒ B (3, 2) \)
Altitude from C is CH \(= 2x + y – 7 = 0 ⇒ C (2, 3) \)
Centroid of ΔABC = E(2, 2) OE = \(2\sqrt2\)
Top Questions on Coordinate Geometry
Let \( F_1, F_2 \) \(\text{ be the foci of the hyperbola}\) \[ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1, a > 0, \, b > 0, \] and let \( O \) be the origin. Let \( M \) be an arbitrary point on curve \( C \) and above the X-axis and \( H \) be a point on \( MF_1 \) such that \( MF_2 \perp F_1 F_2, \, M F_1 \perp OH, \, |OH| = \lambda |O F_2| \) with \( \lambda \in (2/5, 3/5) \), then the range of the eccentricity \( e \) is in:
- NIMCET - 2025
- Mathematics
- Coordinate Geometry
Let the line $\frac{x}{4} + \frac{y}{2} = 1$ meet the x-axis and y-axis at A and B, respectively. M is the midpoint of side AB, and M' is the image of the point M across the line $x + y = 1$. Let the point P lie on the line $x + y = 1$ such that $\Delta ABP$ is an isosceles triangle with $AP = BP$. Then the distance between M' and P is:
- NIMCET - 2025
- Mathematics
- Coordinate Geometry
- The circle \( x^2 + y^2 + ax + by + \gamma = 0 \) is the image of the circle \( x^2 + y^2 - 6x - 10y + 30 = 0 \) across the line \( 3x + y = 2 \). The value of \( [\alpha + \beta + \gamma] \) is (where \( [.] \) represents the floor function)
- NIMCET - 2025
- Mathematics
- Coordinate Geometry
- An equilateral triangle is inscribed in the parabola \( y^2 = x \). One vertex of the triangle is at the vertex of the parabola. The centroid of the triangle is:
- NIMCET - 2025
- Mathematics
- Coordinate Geometry
- A circle with its center in the first quadrant touches both the coordinate axes and the line \( x - y - 2 = 0 \). Then the area of the circle is:
- NIMCET - 2025
- Mathematics
- Coordinate Geometry
Questions Asked in JEE Main exam
- Density of 3 M NaCl solution is 1.25 g/mL. The molality of the solution is:
- JEE Main - 2025
- Solutions
In the given circuit the sliding contact is pulled outwards such that the electric current in the circuit changes at the rate of 8 A/s. At an instant when R is 12 Ω, the value of the current in the circuit will be A.- JEE Main - 2025
- Electrical Circuits
- Let \( \vec{a} = \hat{i} + 2\hat{j} + \hat{k} \) and \( \vec{b} = 2\hat{i} + 7\hat{j} + 3\hat{k} \). Let \( L_1 : \vec{r} = (-\hat{i} + 2\hat{j} + \hat{k}) + \lambda \vec{a}, \lambda \in {R} \) and \( L_2 : \vec{r} = (\hat{j} + \hat{k}) + \mu \vec{b}, \mu \in \mathbb{R} \) be two lines. If the line \( L_3 \) passes through the point of intersection of \( L_1 \) and \( L_2 \), and is parallel to \( \vec{a} + \vec{b} \), then \( L_3 \) passes through the point:
- JEE Main - 2025
- Vectors
Let $ P_n = \alpha^n + \beta^n $, $ n \in \mathbb{N} $. If $ P_{10} = 123,\ P_9 = 76,\ P_8 = 47 $ and $ P_1 = 1 $, then the quadratic equation having roots $ \alpha $ and $ \frac{1}{\beta} $ is:
- JEE Main - 2025
- Relations and functions
For $ \alpha, \beta, \gamma \in \mathbb{R} $, if $$ \lim_{x \to 0} \frac{x^2 \sin \alpha x + (\gamma - 1)e^{x^2} - 3}{\sin 2x - \beta x} = 3, $$ then $ \beta + \gamma - \alpha $ is equal to:
Concepts Used:
Coordinate Geometry
The central idea in coordinate geometry is to assign numerical coordinates to points in a plane or space, which allows us to describe their positions and relationships using algebraic equations. The most common coordinate system is the Cartesian coordinate system, named after the French mathematician and philosopher René Descartes.