If the lines x^2+2xy-35y^2-4x+44y-12=0 and 5x | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) The given equation of the pair of lines is $x^2+2xy-35y^2-4x+44y-12=0$. Comparing this with the general form $ax^2+2hxy+by^2+2gx+2fy+c=0$,we get $

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(C) The given equation of the pair of lines is $x^2+2xy-35y^2-4x+44y-12=0$. Comparing this with the general form $ax^2+2hxy+by^2+2gx+2fy+c=0$,we get $a=1, h=1, b=-35, g=-2, f=22, c=-12$. The point of intersection $(x, y)$ of these lines is given by the formula: $x = \frac{hf-bg}{ab-h^2} = \frac{(1)(22)-(-35)(-2)}{(1)(-35)-(1)^2} = \frac{22-70}{-36} = \frac{-48}{-36} = \frac{4}{3}$ $y = \frac{gh-af}{ab-h^2} = \frac{(-2)(1)-(1)(22)}{(1)(-35)-(1)^2} = \frac{-2-22}{-36} = \frac{-24}{-36} = \frac{2}{3}$ Since the lines are concurrent,the point $(\frac{4}{3}, \frac{2}{3})$ must satisfy the equation $5x+ky-8=0$. Substituting the values: $5(\frac{4}{3}) + k(\frac{2}{3}) - 8 = 0$ $\frac{20}{3} + \frac{2k}{3} = 8$ $20 + 2k = 24$ $2k = 4$ $k = 2$

If the lines $x^2+2xy-35y^2-4x+44y-12=0$ and $5x+ky-8=0$ are concurrent,then $k$ equals

Similar Questions

The solubility of silver bromide in hypo solution is due to the formation of:

Identify the product $(A)$ of the following reaction: $C_2H_5-O-C_2H_5 + CO \xrightarrow[500 \ atm]{BF_3 / 150^{\circ}C} A$

In a certain test there are $n$ questions. In the test,${2^{n - i}}$ students gave wrong answers to at least $i$ questions,where $i = 1, 2, \dots, n$. If the total number of wrong answers given is $2047$,then $n$ is equal to:

The output characteristics of an $n-p-n$ transistor represent,($I_C=$ collector current,$V_{C E}=$ potential difference between collector and emitter,$I_B=$ base current,$V_{B B}=$ voltage given to base,$V_{B E}=$ the potential difference between base and emitter)

Vedclass Test Series

Exam Paper Generator

Online Exam Module