Question:
Two vertices of a triangle are at \(-\hat{i}+3\hat{j}\) and \(2\hat{i}+5\hat{j}\), and its orthocenter is at \(\hat{i}+2\hat{j}\). If the position vector of the \(3^{rd}\) vertex is \(a\hat{i}+b\hat{j}\), then \((a,b)=\)
Two vertices of a triangle are at \(-\hat{i}+3\hat{j}\) and \(2\hat{i}+5\hat{j}\), and its orthocenter is at \(\hat{i}+2\hat{j}\). If the position vector of the \(3^{rd}\) vertex is \(a\hat{i}+b\hat{j}\), then \((a,b)=\)
Show Hint
If \(H\) is the orthocenter of a triangle, then each line joining a vertex to \(H\) is perpendicular to the opposite side.
Updated On: Jun 26, 2026
- \(\left(\dfrac{5}{7},\dfrac{5}{7}\right)\)
- \(\left(\dfrac{5}{7},\dfrac{17}{7}\right)\)
- \(\left(-\dfrac{5}{7},\dfrac{17}{7}\right)\)
- \(\left(\dfrac{5}{7},-\dfrac{17}{7}\right)\)
Show Solution
Verified By Collegedunia
The Correct Option is B
Solution and Explanation
Step 1: Write the given points.
Let the two given vertices be \[ A=(-1,3) \] and \[ B=(2,5). \] Let the third vertex be \[ C=(a,b). \] The orthocenter is \[ H=(1,2). \]
Step 2: Use the property of orthocenter.
Since \(H\) is the orthocenter, the line \(CH\) is perpendicular to \(AB\).
Also, the line \(AH\) is perpendicular to \(BC\).
Step 3: Use \(CH\perp AB\).
The slope of \(AB\) is \[ m_{AB}=\frac{5-3}{2-(-1)} \] \[ =\frac{2}{3}. \] Therefore, the slope of \(CH\) must be \[ -\frac{3}{2}. \] Now, \[ m_{CH}=\frac{b-2}{a-1}. \] So, \[ \frac{b-2}{a-1}=-\frac{3}{2}. \] Cross multiplying, \[ 2(b-2)=-3(a-1) \] \[ 2b-4=-3a+3 \] \[ 3a+2b=7. \]
Step 4: Use \(AH\perp BC\).
The slope of \(AH\) is \[ m_{AH}=\frac{2-3}{1-(-1)} \] \[ =-\frac{1}{2}. \] Therefore, the slope of \(BC\) must be \[ 2. \] Now, \[ m_{BC}=\frac{b-5}{a-2}. \] So, \[ \frac{b-5}{a-2}=2. \] \[ b-5=2a-4 \] \[ b=2a+1. \]
Step 5: Solve the two equations.
We have \[ 3a+2b=7 \] and \[ b=2a+1. \] Substitute \(b=2a+1\) in \(3a+2b=7\): \[ 3a+2(2a+1)=7 \] \[ 3a+4a+2=7 \] \[ 7a=5 \] \[ a=\frac{5}{7}. \] Then, \[ b=2\left(\frac{5}{7}\right)+1 \] \[ =\frac{10}{7}+\frac{7}{7} \] \[ =\frac{17}{7}. \]
Step 6: Final conclusion.
Therefore, \[ (a,b)=\boxed{\left(\frac{5}{7},\frac{17}{7}\right)} \]
Let the two given vertices be \[ A=(-1,3) \] and \[ B=(2,5). \] Let the third vertex be \[ C=(a,b). \] The orthocenter is \[ H=(1,2). \]
Step 2: Use the property of orthocenter.
Since \(H\) is the orthocenter, the line \(CH\) is perpendicular to \(AB\).
Also, the line \(AH\) is perpendicular to \(BC\).
Step 3: Use \(CH\perp AB\).
The slope of \(AB\) is \[ m_{AB}=\frac{5-3}{2-(-1)} \] \[ =\frac{2}{3}. \] Therefore, the slope of \(CH\) must be \[ -\frac{3}{2}. \] Now, \[ m_{CH}=\frac{b-2}{a-1}. \] So, \[ \frac{b-2}{a-1}=-\frac{3}{2}. \] Cross multiplying, \[ 2(b-2)=-3(a-1) \] \[ 2b-4=-3a+3 \] \[ 3a+2b=7. \]
Step 4: Use \(AH\perp BC\).
The slope of \(AH\) is \[ m_{AH}=\frac{2-3}{1-(-1)} \] \[ =-\frac{1}{2}. \] Therefore, the slope of \(BC\) must be \[ 2. \] Now, \[ m_{BC}=\frac{b-5}{a-2}. \] So, \[ \frac{b-5}{a-2}=2. \] \[ b-5=2a-4 \] \[ b=2a+1. \]
Step 5: Solve the two equations.
We have \[ 3a+2b=7 \] and \[ b=2a+1. \] Substitute \(b=2a+1\) in \(3a+2b=7\): \[ 3a+2(2a+1)=7 \] \[ 3a+4a+2=7 \] \[ 7a=5 \] \[ a=\frac{5}{7}. \] Then, \[ b=2\left(\frac{5}{7}\right)+1 \] \[ =\frac{10}{7}+\frac{7}{7} \] \[ =\frac{17}{7}. \]
Step 6: Final conclusion.
Therefore, \[ (a,b)=\boxed{\left(\frac{5}{7},\frac{17}{7}\right)} \]
Was this answer helpful?
0
0
Top AP EAPCET Mathematics Questions
- $D = \left\{x\in\mathbb{R}: f\left(x\right) =\sqrt{\frac{x - \left|x\right|}{x - \left[x\right]}} \text{is defined} \right\}$ and $C$ be the range of the real function $g(x) = \frac{2x}{4 + x^{2}}$. Then $D \cap C$
- $f\left(x\right) = \frac{x}{e^{x}-1} + \frac{x}{2} + 2 \cos^{3} \frac{x}{2} $ on $R-\left\{0\right\} $ is
- Consider the following lists.
List-I List-II (A) $f(x) = \frac{|x+2|}{x+2} , x \ne -2 $ (I) $[\frac{1}{3} , 1 ]$ (B) $(x)=|[x]|,x \in [R$ (II) Z (C) $h(x) = |x - [x]| , x \in [R$ (III) W (D) $f(x) = \frac{1}{2 - \sin 3x} , x \in [R$ (IV) [0, 1) (V) { -1, 1} - If $[x]$ is the greatest integer less than or equal to $x$ and $|x|$ is the modulus of $x$. then the system of three equations $2x + 3 | y | + 5[z] = 0, x + |y| - 2[z] = 4, x + |y| + |z| = 1$ has
- Investigate the values of $\lambda$ and $\mu$ for the system $x+2 y+3 z=6, x+3 y+5 z=9$, $2 x+5 y+\lambda z=\mu$ and match the values in List - I with the items in List - II.
List I List II (A) $\lambda=8, \mu \neq 15$ 1. Infinitely many solutions (B) $\lambda \neq 8, \mu \in R$ 2. No solution (C) $\lambda=8, \mu=15$ 3. Unique solution
View More Questions
Top AP EAPCET Geometry and Vectors Questions
- If \( \vec{i} - 2\vec{j} + 3\vec{k}, 2\vec{i} + 3\vec{j} - \vec{k}, -3\vec{i} - \vec{j} - 2\vec{k} \) are the position vectors of three points A, B, C respectively, then A, B, C:
- If a point \( C \) divides the line segment joining the points with position vectors \( 2\mathbf{i} - 3\mathbf{j} + 2\mathbf{k} \) and \( 3\mathbf{i} - \mathbf{j} - 2\mathbf{k} \) in the ratio \( 2:3 \), then the distance of \( C \) from the point with position vector \( 2\mathbf{i} - \mathbf{j} + \mathbf{k} \) is
- If \( \mathbf{a} = (2x + y)\mathbf{i} + 3\mathbf{j} + 9\mathbf{k} \) and \( \mathbf{b} = 2\mathbf{i} + \mathbf{j} - (x - y)\mathbf{k} \) are two collinear vectors, then \( x^3 + 27y^3 = \)
- Let \( \mathbf{a} = \mathbf{i} - 2\mathbf{j} \), \( \mathbf{b} = 2\mathbf{j} + 3\mathbf{k} \), \( \mathbf{c} = p\mathbf{i} + q\mathbf{j} \) and \( \mathbf{d} = p\mathbf{j} - q\mathbf{k} \) be four vectors. If \( (\mathbf{a} \times \mathbf{b}) \cdot \mathbf{c} = 3 \) and \( (\mathbf{a} \times \mathbf{b}) \cdot \mathbf{d} = 0 \), then \( 3p + q = \)
- Let \( \mathbf{a} = \mathbf{i} + 2\mathbf{j} + 3\mathbf{k} \) and \( \mathbf{b} = \mathbf{i} - 2\mathbf{j} - 3\mathbf{k} \) be two vectors. If \( A_1 \) is the area of the quadrilateral having \( \mathbf{a}, \mathbf{b} \) as its diagonals and \( A_2 \) is the area of the parallelogram having \( \mathbf{a}, \mathbf{b} \) as its adjacent sides, then \( A_1 : A_2 = \)
View More Questions
Top AP EAPCET Questions
- A body is projected from the ground at an angle of $\tan^{-1} \left( \frac{8}{7}\right)$ with the horizontal. The ratio of the maximum height attained by it to its range is
- The four quantum numbers of the valence electron of potassium are :
- A lens forms real and virtual images of an object, when the object is at $u_{1}$ and $u_{2}$ distances respectively. If the size of the virtual image is double that of the real image, then the focal length of the lens is (take, the magnification of the real image as $m$ )
- Two spheres $P$ and $Q$, each of mass $200\, g$ are attached to a string of length one metre as shown in the figure. The string and the spheres are then whirled in a horizontal circle about $O$ at a constant angular speed. The ratio of the tension in the string between $P$ and $Q$ to that of between $P$ and $O$ is $(P$ is at mid-point of the line joining $O$ and $Q$ )
- The volume of $0.02\, M$ acidified permanganate solution required for complete reaction of $60\, mL$ of $0.01\, MI^{-}$ ion solution to form $I_2$ in $mL$ is
View More Questions