Question:
The product of the perpendiculars from $(-1, 2)$ to the pair of lines $2x^{2} - 5xy + 2y^{2} + 3x - 3y + 1 = 0$ is
The product of the perpendiculars from $(-1, 2)$ to the pair of lines $2x^{2} - 5xy + 2y^{2} + 3x - 3y + 1 = 0$ is
Show Hint
Product of perpendicular distances from $(x_1,y_1)$ to the pair of lines $ax^2+2hxy+by^2+\cdots = 0$ equals $\dfrac{|ax_1^2+2hx_1y_1+by_1^2+2gx_1+2fy_1+c|}{\sqrt{(a-b)^2+4h^2}}$.
Updated On: May 2, 2026
- $\dfrac{5}{12}$
- $\dfrac{12}{5}$
- $\dfrac{6}{5}$
- $\dfrac{5}{6}$
Show Solution
Verified By Collegedunia
The Correct Option is B
Solution and Explanation
Step 1: Understanding the Concept:
For a pair of lines $ax^2+2hxy+by^2+2gx+2fy+c=0$, the product of perpendicular distances from $(x_1,y_1)$ is given by a standard formula.
Step 2: Detailed Explanation:
Here $a=2$, $h=-5/2$, $b=2$, $g=3/2$, $f=-3/2$, $c=1$.
Numerator $= |ax_1^2 + 2hx_1y_1 + by_1^2 + 2gx_1 + 2fy_1 + c|$ evaluated at $(-1,2)$.
$= |2(1) + 2(-5/2)(-1)(2) + 2(4) + 2(3/2)(-1) + 2(-3/2)(2) + 1| = |2 + 10 + 8 - 3 - 6 + 1| = 12$.
Denominator $= \sqrt{(a-b)^2 + 4h^2} = \sqrt{0 + 25} = 5$.
Product $= \dfrac{12}{5}$.
Step 3: Final Answer:
The product of perpendiculars is $\dfrac{12}{5}$.
For a pair of lines $ax^2+2hxy+by^2+2gx+2fy+c=0$, the product of perpendicular distances from $(x_1,y_1)$ is given by a standard formula.
Step 2: Detailed Explanation:
Here $a=2$, $h=-5/2$, $b=2$, $g=3/2$, $f=-3/2$, $c=1$.
Numerator $= |ax_1^2 + 2hx_1y_1 + by_1^2 + 2gx_1 + 2fy_1 + c|$ evaluated at $(-1,2)$.
$= |2(1) + 2(-5/2)(-1)(2) + 2(4) + 2(3/2)(-1) + 2(-3/2)(2) + 1| = |2 + 10 + 8 - 3 - 6 + 1| = 12$.
Denominator $= \sqrt{(a-b)^2 + 4h^2} = \sqrt{0 + 25} = 5$.
Product $= \dfrac{12}{5}$.
Step 3: Final Answer:
The product of perpendiculars is $\dfrac{12}{5}$.
Was this answer helpful?
0
0
Top MET Mathematics Questions
- $\int\limits_{0}^{8}| x -5| d x$ is equal to
- The equation $3^{3x + 4} = 9^{2x - 2},\ x>0$ has the solution
- A parallelogram is constructed on the vectors $\mathbf{a} = 3\alpha - \beta$, $\mathbf{b} = \alpha + 3\beta$. If $|\alpha| = |\beta| = 2$ and the angle between $\alpha$ and $\beta$ is $\dfrac{\pi}{3}$, then the length of a diagonal of the parallelogram is
- Every group of order 7 is
- Let $U_{n} = 2 + 2^{3} + 2^{5} + \cdots + 2^{2n+1}$ and $V_{n} = 1 + 4 + 4^{2} + \cdots + 4^{n-1}$. Then $\displaystyle\lim_{n \to \infty} \dfrac{U_n}{V_n}$ is equal to
View More Questions
Top MET Straight lines Questions
- $A(-1,2)$, $B(5,1)$, $C(6,5)$ are the vertices of a parallelogram $ABCD$. The equation of the diagonal through $B$ is
- The line $\frac{x}{a} - \frac{y}{b} = 1$ cuts the x-axis at P. The equation of the line through P perpendicular to the given line is
- If the angle between two lines represented by $2x^2 + 5xy + 3y^2 + 7y + 4 = 0$ is $\tan^{-1} m$, then m is equal to
- The foot of the perpendicular from $(-2,3)$ to the line $2x - y - 3 = 0$ is
- The equation of the line passing through the point $(1, 2)$ and perpendicular to the line $x + y + 1 = 0$ is:
View More Questions
Top MET Questions
- A coil of 100 turns and area $2 \times 10^{-2}m^2 $ is pivoted about a vertical diameter in a uniform magnetic field and carries a current of 5A. When the coil is held with its plane in north-south direction, it experiences a couple of 0.33 Nm. When the plane is east-west, the corresponding couple is 0.4 Nm, the value of magnetic induction is [Neglect earth's magnetic field]
- $\int\limits_{0}^{8}| x -5| d x$ is equal to
- Radiation, with wavelength 6561 $?$ falls on a metal surface to produce photoelectrons. The electrons are made to enter a uniform magnetic field of $3 \times 10^{-4} T$. If the radius of the largest circular path followed by the electrons is 10 mm, the work function of the metal is close to :
- In intrinsic semiconductor at room temperature number of electrons and holes are
- Which of the following product is formed in the reaction $ CH_3MgBr {->[(i)CO_2][(ii)H_2O]} ? $
View More Questions