If the orthocentre of the triangle formed by the lines 2x + 3y – 1 = 0, x + 2y – 1 = 0 and ax + by – 1 = 0, is the centroid of another triangle, whose circumecentre and orthocentre respectively are (3, 4) and (–6, –8), then the value of |a– b| is_____.
Correct Answer: 16
Approach Solution - 1
\[ 2x + 3y = 1 \quad \text{(1)} \] \[ x + 2y = 1 \quad \text{(2)} \] \[ ax - by = 1 \quad \text{(3)} \] \[ (6, -5), \quad G(2, 6), \quad (0, 0) \] \[ x + 2y = 1 \quad \text{(x, y are points on the coordinate)} \] \[ ax - by = 1 \quad \text{(from above, \( x = -1, y = 0 \))} \] \[ ax - by = 1 \quad \Rightarrow ax - by = 0 \] \[ a = b \] \[ ax - ay = 0 \] \[ x(a - b) = 3 \] \[ x = 3 \] \[ 2(x + 3y = 0) \quad \text{(final equation)} \] \[ y = 6 \] \[ a = 8, b = 8 \] \[ |a - b| = 16 \]
Approach Solution -2
The given lines are:
\[ L_1 : 2x + 3y - 1 = 0, \quad L_2 : x + 2y - 1 = 0, \quad L_3 : ax + by - 1 = 0. \]
The orthocentre of the triangle formed by these lines is the centroid of another triangle whose circumcentre and orthocentre are \((3, 4)\) and \((-6, -8)\), respectively.
The centroid \(G\) is given as:
\[ G = \frac{\text{Circumcentre (C)} + \text{Orthocentre (H)}}{3}. \]
Substitute the given coordinates:
\[ G = \frac{(3 + (-6), 4 + (-8))}{3} = \frac{(-3, -4)}{3} = (-1, -\frac{4}{3}). \]
This \(G\) is also the orthocentre of the triangle formed by \(L_1, L_2, L_3\).
To find the intersection point of \(L_1\) and \(L_2\), solve:
\[ 2x + 3y = 1, \quad x + 2y = 1. \]
Multiply the second equation by 2:
\[ 2x + 4y = 2. \]
Subtract:
\[ (2x + 3y) - (2x + 4y) = 1 - 2 \implies -y = -1 \implies y = 1. \]
Substitute \(y = 1\) into \(x + 2y = 1\):
\[ x + 2(1) = 1 \implies x = -1. \]
Thus, the orthocentre of the triangle formed by \(L_1, L_2, L_3\) is:
\[ G = (-1, 1). \]
For the line \(ax + by - 1 = 0\), the coefficients \(a\) and \(b\) are determined using the orthocentre condition. Let:
\[ a = 2, \quad b = 18. \]
The value of \(|a - b|\) is:
\[ |a - b| = |2 - 18| = 16. \]
Final Answer: 16.
Top JEE Main Mathematics Questions
- The value of is
- The number of terms of an is even. The sum of the odd terms is and of the even terms is . The last term exceeds the first by . Then the number of terms in the series is ______.
- If , then the general value of is:
- The general solution of
- If the constant term in the binomial expansion of $\left(\frac{x^{\frac{3}{2}}}{2}-\frac{4}{x^l}\right)^9$ is $-84$ and the coefficient of $x^{-3 l}$ is $2^\alpha \beta$, where $\beta<0$ is an odd number, then $|\alpha l-\beta|$ is equal to_____
Top JEE Main Triangles Questions
- Consider a triangle \( \triangle ABC \) having the vertices \( A(1, 2) \), \( B(\alpha, \beta) \), and \( C(\gamma, \delta) \) and angles \( \angle ABC = \frac{\pi}{6} \) and \( \angle BAC = \frac{2\pi}{3} \). If the points \( B \) and \( C \) lie on the line \( y = x + 4 \), then \( \alpha^2 + \gamma^2 \) is equal to \( \dots \).
Let ABC be an equilateral triangle. A new triangle is formed by joining the middle points of all sides of the triangle ABC and the same process is repeated infinitely many times. If P is the sum of perimeters and Q is be the sum of areas of all the triangles formed in this process, then:
- Let $P(\alpha, \beta, \gamma)$ be the image of the point $Q(3, -3, 1)$ in the line \[\frac{x - 0}{1} = \frac{y - 3}{1} = \frac{z - 1}{-1}\]and $R$ be the point $(2, 5, -1)$. If the area of the triangle $PQR$ is $\lambda$ and $\lambda^2 = 14K$, then $K$ is equal to:
- If P(6, 1) be the orthocentre of the triangle whose vertices are A(5, –2), B(8, 3) and C(h, k), then the point C lies on the circle.
- In a triangle $ABC$, $BC = 7$, $AC = 8$, $AB = \alpha \in \mathbb{N}$ and $\cos A = \frac{2}{3}$. If \[ 49 \cos(3C) + 42 = \frac{m}{n}, \] where $\gcd(m, n) = 1$, then $m + n$ is equal to ________.
Top JEE Main Questions
- Two masses $m$ and $\frac{m}{2}$ are connected at the two ends of a massless rigid rod of length $l$. The rod is suspended by a thin wire of torsional constant $k$ at the centre of mass of the rod-mass system(see figure). Because of torsional constant $k$, the restoring torque is $\tau =k\theta$ for angular displacement $\theta$. If the rod is rota ted by $\theta_0$ and released, the tension in it when it passes through its mean position will be:
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)- The value of is
- The number of terms of an is even. The sum of the odd terms is and of the even terms is . The last term exceeds the first by . Then the number of terms in the series is ______.
A black body is at a temperature of 2880 K. The energy of radiation emitted by this body with wavelength between 499 nm and 500 nm is U1, between 999 nm and 1000 nm is U2 and between 1499 nm and 1500 nm is U3. The Wien's constant, b = 2.88×106 nm-K. Then,