88419
If the equation \(k x y+5 x+3 y+2=0\) represents a pair of lines, then \(k=\)
1 \(0, \frac{-15}{2}\)
2 15,1
3 \(0, \frac{15}{2}\)
4 \(1, \frac{15}{2}\)
Explanation:
(C) : The given equation is \(\mathrm{kxy}+5 \mathrm{x}+3 \mathrm{y}+2=0\) Comparing given equation with \(a x^{2}+2 h x y+b y^{2}+2 g x+2 f y+c=0\), we get \(\mathrm{a}=0, \mathrm{~b}=0, \mathrm{c}=2, \mathrm{f}=\frac{3}{2}, \mathrm{~g}=\frac{5}{2}, \mathrm{~h}=\frac{\mathrm{k}}{2}\) \(\Delta=\mathrm{abc}+2 \mathrm{fgh}-\mathrm{af}^{2}-\mathrm{bg}^{2}-\mathrm{ch}^{2}=0\) \(\Delta=0+2 \frac{(3)}{2} \frac{(5)}{2}\left(\frac{\mathrm{k}}{2}\right)-0-0-2\left(\frac{\mathrm{k}}{2}\right)^{2}=0\) \(\frac{15}{4} \mathrm{k}-\frac{\mathrm{k}^{2}}{2}=0 \Rightarrow 15 \mathrm{k}-2 \mathrm{k}^{2}-0 \Rightarrow \mathrm{k}(2 \mathrm{k}-15)=0 \Rightarrow \mathrm{k}=0, \frac{15}{2}\)
[ -2020]
Co-Ordinate system
88420
The separate equations of the lines represented by the equation \(3 x^{2}-2 \sqrt{3} x y-3 y^{2}=0\) are
1 \(x+\sqrt{3} y=0\) and \(3 \mathrm{x}-\sqrt{3} \mathrm{y}=0\)
2 \(x+\sqrt{3} y=0\) and \(\quad 3 x+\sqrt{3} y=0\)
3 \(x-\sqrt{3} y=0\) and \(3 \mathrm{x}-\sqrt{3} \mathrm{y}=0\)
4 \(x-\sqrt{3} y=0\) and \(\quad 3 x+\sqrt{3} y=0\)
Explanation:
(D) : Given the equation, \(3 x^{2}-2 \sqrt{3} x y-3 y^{2}=0\) \(3 x^{2}-3 \sqrt{3} x y-\sqrt{3} x y-3 y^{2}-0\) \(3 \mathrm{x}(\mathrm{x}-\sqrt{3} \mathrm{y})+\sqrt{3} \mathrm{y}(\mathrm{x}-\sqrt{3} \mathrm{y})=0\) \((3 x+\sqrt{3} y)(x-\sqrt{3} y)=0\) The separate equation of the lines are \(3 x+\sqrt{3} y=0\) and \(x-\sqrt{3} y=0\)
MHT CET-2020
Co-Ordinate system
88421
The equation of a line passing through the point of intersection of the lines \(x+2 y+8=0\) and \(3 x-y+4=0\) having \(x\) and \(y\) intercept zero is
1 \(4 x+5 y=0\)
2 \(4 x-5 y=0\)
3 \(5 x+4 y=0\)
4 \(5 x-4 y=0\)
Explanation:
(D) :Given, \(\mathrm{x}+2 \mathrm{y}+8=0\) \(3 \mathrm{x}-\mathrm{y}+4=0\) By solving (1) and (2) we get \(x=\frac{-16}{7}, y=\frac{-20}{7}\) Since required line does not have intercept on \(\mathrm{X}\) and \(\mathrm{Y}\) axis, it passes through origin. Hence required equation is \(\frac{y+\frac{20}{7}}{\frac{20}{7}}=\frac{x+\frac{16}{7}}{\frac{16}{7}} \Rightarrow \frac{7 y+20}{20}=\frac{7 x+16}{16} \Rightarrow 5 x-4 y=0\)
MHT CET-2020
Co-Ordinate system
88422
The joint equation of a pair of lines passing through \((2,3)\) and parallel to the lines \(x^{2}-y^{2}=\) 0 is
1 \(x^{2}-y^{2}-4 x+6 y+2=0\)
2 \(x^{2}-y^{2}-4 x+6 y=0\)
3 \(x^{2}-y^{2}-4 x+6 y-5=0\)
4 \(x^{2}-y^{2}-4 x+6 y+17=0\)
Explanation:
(C) : Given, line \(\mathrm{x}^{2}-\mathrm{y}^{2}=0 \Rightarrow(\mathrm{x}-\mathrm{y})(\mathrm{x}+\mathrm{y})=0\) Thus slopes of given lines are 1 and -1 . The equation of the line passing through \((2,3)\) and parallel to given lines are \((y-3)=(x-2)\) and \((y-3)=-(x-2)\) \(\mathrm{x}-\mathrm{y}+1=0\) and \((\mathrm{x}+\mathrm{y}-5)=0\) Hence required equation is \((x-y+1)(x+y-5)=0\) \(x^{2}-x y+x+x y-y^{2}+y-5 x+5 y=0\) \(x^{2}-y^{2}-4 x+6 y-5=0\)
NEET Test Series from KOTA - 10 Papers In MS WORD
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Co-Ordinate system
88419
If the equation \(k x y+5 x+3 y+2=0\) represents a pair of lines, then \(k=\)
1 \(0, \frac{-15}{2}\)
2 15,1
3 \(0, \frac{15}{2}\)
4 \(1, \frac{15}{2}\)
Explanation:
(C) : The given equation is \(\mathrm{kxy}+5 \mathrm{x}+3 \mathrm{y}+2=0\) Comparing given equation with \(a x^{2}+2 h x y+b y^{2}+2 g x+2 f y+c=0\), we get \(\mathrm{a}=0, \mathrm{~b}=0, \mathrm{c}=2, \mathrm{f}=\frac{3}{2}, \mathrm{~g}=\frac{5}{2}, \mathrm{~h}=\frac{\mathrm{k}}{2}\) \(\Delta=\mathrm{abc}+2 \mathrm{fgh}-\mathrm{af}^{2}-\mathrm{bg}^{2}-\mathrm{ch}^{2}=0\) \(\Delta=0+2 \frac{(3)}{2} \frac{(5)}{2}\left(\frac{\mathrm{k}}{2}\right)-0-0-2\left(\frac{\mathrm{k}}{2}\right)^{2}=0\) \(\frac{15}{4} \mathrm{k}-\frac{\mathrm{k}^{2}}{2}=0 \Rightarrow 15 \mathrm{k}-2 \mathrm{k}^{2}-0 \Rightarrow \mathrm{k}(2 \mathrm{k}-15)=0 \Rightarrow \mathrm{k}=0, \frac{15}{2}\)
[ -2020]
Co-Ordinate system
88420
The separate equations of the lines represented by the equation \(3 x^{2}-2 \sqrt{3} x y-3 y^{2}=0\) are
1 \(x+\sqrt{3} y=0\) and \(3 \mathrm{x}-\sqrt{3} \mathrm{y}=0\)
2 \(x+\sqrt{3} y=0\) and \(\quad 3 x+\sqrt{3} y=0\)
3 \(x-\sqrt{3} y=0\) and \(3 \mathrm{x}-\sqrt{3} \mathrm{y}=0\)
4 \(x-\sqrt{3} y=0\) and \(\quad 3 x+\sqrt{3} y=0\)
Explanation:
(D) : Given the equation, \(3 x^{2}-2 \sqrt{3} x y-3 y^{2}=0\) \(3 x^{2}-3 \sqrt{3} x y-\sqrt{3} x y-3 y^{2}-0\) \(3 \mathrm{x}(\mathrm{x}-\sqrt{3} \mathrm{y})+\sqrt{3} \mathrm{y}(\mathrm{x}-\sqrt{3} \mathrm{y})=0\) \((3 x+\sqrt{3} y)(x-\sqrt{3} y)=0\) The separate equation of the lines are \(3 x+\sqrt{3} y=0\) and \(x-\sqrt{3} y=0\)
MHT CET-2020
Co-Ordinate system
88421
The equation of a line passing through the point of intersection of the lines \(x+2 y+8=0\) and \(3 x-y+4=0\) having \(x\) and \(y\) intercept zero is
1 \(4 x+5 y=0\)
2 \(4 x-5 y=0\)
3 \(5 x+4 y=0\)
4 \(5 x-4 y=0\)
Explanation:
(D) :Given, \(\mathrm{x}+2 \mathrm{y}+8=0\) \(3 \mathrm{x}-\mathrm{y}+4=0\) By solving (1) and (2) we get \(x=\frac{-16}{7}, y=\frac{-20}{7}\) Since required line does not have intercept on \(\mathrm{X}\) and \(\mathrm{Y}\) axis, it passes through origin. Hence required equation is \(\frac{y+\frac{20}{7}}{\frac{20}{7}}=\frac{x+\frac{16}{7}}{\frac{16}{7}} \Rightarrow \frac{7 y+20}{20}=\frac{7 x+16}{16} \Rightarrow 5 x-4 y=0\)
MHT CET-2020
Co-Ordinate system
88422
The joint equation of a pair of lines passing through \((2,3)\) and parallel to the lines \(x^{2}-y^{2}=\) 0 is
1 \(x^{2}-y^{2}-4 x+6 y+2=0\)
2 \(x^{2}-y^{2}-4 x+6 y=0\)
3 \(x^{2}-y^{2}-4 x+6 y-5=0\)
4 \(x^{2}-y^{2}-4 x+6 y+17=0\)
Explanation:
(C) : Given, line \(\mathrm{x}^{2}-\mathrm{y}^{2}=0 \Rightarrow(\mathrm{x}-\mathrm{y})(\mathrm{x}+\mathrm{y})=0\) Thus slopes of given lines are 1 and -1 . The equation of the line passing through \((2,3)\) and parallel to given lines are \((y-3)=(x-2)\) and \((y-3)=-(x-2)\) \(\mathrm{x}-\mathrm{y}+1=0\) and \((\mathrm{x}+\mathrm{y}-5)=0\) Hence required equation is \((x-y+1)(x+y-5)=0\) \(x^{2}-x y+x+x y-y^{2}+y-5 x+5 y=0\) \(x^{2}-y^{2}-4 x+6 y-5=0\)
88419
If the equation \(k x y+5 x+3 y+2=0\) represents a pair of lines, then \(k=\)
1 \(0, \frac{-15}{2}\)
2 15,1
3 \(0, \frac{15}{2}\)
4 \(1, \frac{15}{2}\)
Explanation:
(C) : The given equation is \(\mathrm{kxy}+5 \mathrm{x}+3 \mathrm{y}+2=0\) Comparing given equation with \(a x^{2}+2 h x y+b y^{2}+2 g x+2 f y+c=0\), we get \(\mathrm{a}=0, \mathrm{~b}=0, \mathrm{c}=2, \mathrm{f}=\frac{3}{2}, \mathrm{~g}=\frac{5}{2}, \mathrm{~h}=\frac{\mathrm{k}}{2}\) \(\Delta=\mathrm{abc}+2 \mathrm{fgh}-\mathrm{af}^{2}-\mathrm{bg}^{2}-\mathrm{ch}^{2}=0\) \(\Delta=0+2 \frac{(3)}{2} \frac{(5)}{2}\left(\frac{\mathrm{k}}{2}\right)-0-0-2\left(\frac{\mathrm{k}}{2}\right)^{2}=0\) \(\frac{15}{4} \mathrm{k}-\frac{\mathrm{k}^{2}}{2}=0 \Rightarrow 15 \mathrm{k}-2 \mathrm{k}^{2}-0 \Rightarrow \mathrm{k}(2 \mathrm{k}-15)=0 \Rightarrow \mathrm{k}=0, \frac{15}{2}\)
[ -2020]
Co-Ordinate system
88420
The separate equations of the lines represented by the equation \(3 x^{2}-2 \sqrt{3} x y-3 y^{2}=0\) are
1 \(x+\sqrt{3} y=0\) and \(3 \mathrm{x}-\sqrt{3} \mathrm{y}=0\)
2 \(x+\sqrt{3} y=0\) and \(\quad 3 x+\sqrt{3} y=0\)
3 \(x-\sqrt{3} y=0\) and \(3 \mathrm{x}-\sqrt{3} \mathrm{y}=0\)
4 \(x-\sqrt{3} y=0\) and \(\quad 3 x+\sqrt{3} y=0\)
Explanation:
(D) : Given the equation, \(3 x^{2}-2 \sqrt{3} x y-3 y^{2}=0\) \(3 x^{2}-3 \sqrt{3} x y-\sqrt{3} x y-3 y^{2}-0\) \(3 \mathrm{x}(\mathrm{x}-\sqrt{3} \mathrm{y})+\sqrt{3} \mathrm{y}(\mathrm{x}-\sqrt{3} \mathrm{y})=0\) \((3 x+\sqrt{3} y)(x-\sqrt{3} y)=0\) The separate equation of the lines are \(3 x+\sqrt{3} y=0\) and \(x-\sqrt{3} y=0\)
MHT CET-2020
Co-Ordinate system
88421
The equation of a line passing through the point of intersection of the lines \(x+2 y+8=0\) and \(3 x-y+4=0\) having \(x\) and \(y\) intercept zero is
1 \(4 x+5 y=0\)
2 \(4 x-5 y=0\)
3 \(5 x+4 y=0\)
4 \(5 x-4 y=0\)
Explanation:
(D) :Given, \(\mathrm{x}+2 \mathrm{y}+8=0\) \(3 \mathrm{x}-\mathrm{y}+4=0\) By solving (1) and (2) we get \(x=\frac{-16}{7}, y=\frac{-20}{7}\) Since required line does not have intercept on \(\mathrm{X}\) and \(\mathrm{Y}\) axis, it passes through origin. Hence required equation is \(\frac{y+\frac{20}{7}}{\frac{20}{7}}=\frac{x+\frac{16}{7}}{\frac{16}{7}} \Rightarrow \frac{7 y+20}{20}=\frac{7 x+16}{16} \Rightarrow 5 x-4 y=0\)
MHT CET-2020
Co-Ordinate system
88422
The joint equation of a pair of lines passing through \((2,3)\) and parallel to the lines \(x^{2}-y^{2}=\) 0 is
1 \(x^{2}-y^{2}-4 x+6 y+2=0\)
2 \(x^{2}-y^{2}-4 x+6 y=0\)
3 \(x^{2}-y^{2}-4 x+6 y-5=0\)
4 \(x^{2}-y^{2}-4 x+6 y+17=0\)
Explanation:
(C) : Given, line \(\mathrm{x}^{2}-\mathrm{y}^{2}=0 \Rightarrow(\mathrm{x}-\mathrm{y})(\mathrm{x}+\mathrm{y})=0\) Thus slopes of given lines are 1 and -1 . The equation of the line passing through \((2,3)\) and parallel to given lines are \((y-3)=(x-2)\) and \((y-3)=-(x-2)\) \(\mathrm{x}-\mathrm{y}+1=0\) and \((\mathrm{x}+\mathrm{y}-5)=0\) Hence required equation is \((x-y+1)(x+y-5)=0\) \(x^{2}-x y+x+x y-y^{2}+y-5 x+5 y=0\) \(x^{2}-y^{2}-4 x+6 y-5=0\)
88419
If the equation \(k x y+5 x+3 y+2=0\) represents a pair of lines, then \(k=\)
1 \(0, \frac{-15}{2}\)
2 15,1
3 \(0, \frac{15}{2}\)
4 \(1, \frac{15}{2}\)
Explanation:
(C) : The given equation is \(\mathrm{kxy}+5 \mathrm{x}+3 \mathrm{y}+2=0\) Comparing given equation with \(a x^{2}+2 h x y+b y^{2}+2 g x+2 f y+c=0\), we get \(\mathrm{a}=0, \mathrm{~b}=0, \mathrm{c}=2, \mathrm{f}=\frac{3}{2}, \mathrm{~g}=\frac{5}{2}, \mathrm{~h}=\frac{\mathrm{k}}{2}\) \(\Delta=\mathrm{abc}+2 \mathrm{fgh}-\mathrm{af}^{2}-\mathrm{bg}^{2}-\mathrm{ch}^{2}=0\) \(\Delta=0+2 \frac{(3)}{2} \frac{(5)}{2}\left(\frac{\mathrm{k}}{2}\right)-0-0-2\left(\frac{\mathrm{k}}{2}\right)^{2}=0\) \(\frac{15}{4} \mathrm{k}-\frac{\mathrm{k}^{2}}{2}=0 \Rightarrow 15 \mathrm{k}-2 \mathrm{k}^{2}-0 \Rightarrow \mathrm{k}(2 \mathrm{k}-15)=0 \Rightarrow \mathrm{k}=0, \frac{15}{2}\)
[ -2020]
Co-Ordinate system
88420
The separate equations of the lines represented by the equation \(3 x^{2}-2 \sqrt{3} x y-3 y^{2}=0\) are
1 \(x+\sqrt{3} y=0\) and \(3 \mathrm{x}-\sqrt{3} \mathrm{y}=0\)
2 \(x+\sqrt{3} y=0\) and \(\quad 3 x+\sqrt{3} y=0\)
3 \(x-\sqrt{3} y=0\) and \(3 \mathrm{x}-\sqrt{3} \mathrm{y}=0\)
4 \(x-\sqrt{3} y=0\) and \(\quad 3 x+\sqrt{3} y=0\)
Explanation:
(D) : Given the equation, \(3 x^{2}-2 \sqrt{3} x y-3 y^{2}=0\) \(3 x^{2}-3 \sqrt{3} x y-\sqrt{3} x y-3 y^{2}-0\) \(3 \mathrm{x}(\mathrm{x}-\sqrt{3} \mathrm{y})+\sqrt{3} \mathrm{y}(\mathrm{x}-\sqrt{3} \mathrm{y})=0\) \((3 x+\sqrt{3} y)(x-\sqrt{3} y)=0\) The separate equation of the lines are \(3 x+\sqrt{3} y=0\) and \(x-\sqrt{3} y=0\)
MHT CET-2020
Co-Ordinate system
88421
The equation of a line passing through the point of intersection of the lines \(x+2 y+8=0\) and \(3 x-y+4=0\) having \(x\) and \(y\) intercept zero is
1 \(4 x+5 y=0\)
2 \(4 x-5 y=0\)
3 \(5 x+4 y=0\)
4 \(5 x-4 y=0\)
Explanation:
(D) :Given, \(\mathrm{x}+2 \mathrm{y}+8=0\) \(3 \mathrm{x}-\mathrm{y}+4=0\) By solving (1) and (2) we get \(x=\frac{-16}{7}, y=\frac{-20}{7}\) Since required line does not have intercept on \(\mathrm{X}\) and \(\mathrm{Y}\) axis, it passes through origin. Hence required equation is \(\frac{y+\frac{20}{7}}{\frac{20}{7}}=\frac{x+\frac{16}{7}}{\frac{16}{7}} \Rightarrow \frac{7 y+20}{20}=\frac{7 x+16}{16} \Rightarrow 5 x-4 y=0\)
MHT CET-2020
Co-Ordinate system
88422
The joint equation of a pair of lines passing through \((2,3)\) and parallel to the lines \(x^{2}-y^{2}=\) 0 is
1 \(x^{2}-y^{2}-4 x+6 y+2=0\)
2 \(x^{2}-y^{2}-4 x+6 y=0\)
3 \(x^{2}-y^{2}-4 x+6 y-5=0\)
4 \(x^{2}-y^{2}-4 x+6 y+17=0\)
Explanation:
(C) : Given, line \(\mathrm{x}^{2}-\mathrm{y}^{2}=0 \Rightarrow(\mathrm{x}-\mathrm{y})(\mathrm{x}+\mathrm{y})=0\) Thus slopes of given lines are 1 and -1 . The equation of the line passing through \((2,3)\) and parallel to given lines are \((y-3)=(x-2)\) and \((y-3)=-(x-2)\) \(\mathrm{x}-\mathrm{y}+1=0\) and \((\mathrm{x}+\mathrm{y}-5)=0\) Hence required equation is \((x-y+1)(x+y-5)=0\) \(x^{2}-x y+x+x y-y^{2}+y-5 x+5 y=0\) \(x^{2}-y^{2}-4 x+6 y-5=0\)