NEET Test Series from KOTA - 10 Papers In MS WORD
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Co-Ordinate system
88382
Suppose a point \(P\) moves so that \(\mathbf{B P}^{2}-\mathbf{A P}^{2}=\) 121 , where \(A\) and \(B\) are \((2,5)\) and \((5,11)\) respectively. Then the locus of \(P\) is a straight line, whose slope is
1 \(1 / 2\)
2 -2
3 \(-1 / 2\)
4 2
Explanation:
(C) : Let, coordinate of \(\mathrm{P}\) be \((\mathrm{x}, \mathrm{y})\) \((\mathrm{BP})^{2}-(\mathrm{AP})^{2}=121\) \(\Rightarrow\left[(5-\mathrm{x})^{2}+(11-\mathrm{y})^{2}\right]-\left[(2-\mathrm{x})^{2}+(5-\mathrm{y})^{2}\right]=121\) \(\Rightarrow 25+\mathrm{x}^{2}-10 \mathrm{x}+121+\mathrm{y}^{2}-22 \mathrm{y}-4-\mathrm{x}^{2}+4 \mathrm{x}-25-\) \(\mathrm{y}^{2}+10 \mathrm{y}=121\) \(=-10 x-22 y-4+4 x+10 y=0\) \(-6 x-12 y-4=0\) \(-2(3 x+6 y+2)=0\) \(3 x+6 y+2=0\) Slope of the line is \(=-\frac{\text { coefficient of } x}{\text { coefficient } x^{2}}=\frac{-3}{6}=\frac{-1}{2}\)
AP EAMCET-2022-05.07.2022
Co-Ordinate system
88383
A value of \(\mathrm{k}\) such that the straight lines \(\mathrm{y}-3 \mathrm{kx}\) \(+4=0\) and \((2 k-1) x-(8 k-1) y-6=0\) are perpendicular is
88384
If \(m_{1}, m_{2}\left(m_{1}>m_{2}\right)\) are the slopes of the lines which make an angle of \(30^{\circ}\) with the line joining the points \((1,2)\) and \((3,4)\), then \(\frac{m_{1}}{m_{2}}=\)
88385
In an isosceles right angled triangle, if the equation of the hypotenuse and its opposite vertex are \(3 x+4 y=4\) and \((2,2)\), then the slopes of the remaining two sides are.
1 \(\frac{1}{7},-7\)
2 \(\frac{-1}{7}, 7\)
3 \(\frac{1}{7}, 7\)
4 \(\frac{-1}{7},-7\)
Explanation:
(A): Hypotenuse is along the line \(3 x+4 y-4=0\) \(\therefore \quad\) Slope of \(\mathrm{AC}=\frac{-3}{4}\) \(\because \quad \Delta \mathrm{ABC}\) is isosceles right angled triangle, Now, let the slope of the line making an angle \(45^{\circ}\) with \(\mathrm{AC}\) be \(\mathrm{m}-\) \(\tan 45^{\circ}=\left|\frac{m-\left(\frac{-3}{4}\right)}{1+m\left(\frac{-3}{4}\right)}\right|\) \(\therefore\) \(\pm 1=\frac{4 m+3}{4-3 m}\) \(4 m+3=4-3 m \quad \text { or } \quad 4 m+3=-4+3 m\) \(7 m=1\) \(m=\frac{1}{7}\) So, slope of remaining two sides \(\mathrm{BC}\) and \(\mathrm{AB}\) are \(\frac{1}{7}\) and -7 .
88382
Suppose a point \(P\) moves so that \(\mathbf{B P}^{2}-\mathbf{A P}^{2}=\) 121 , where \(A\) and \(B\) are \((2,5)\) and \((5,11)\) respectively. Then the locus of \(P\) is a straight line, whose slope is
1 \(1 / 2\)
2 -2
3 \(-1 / 2\)
4 2
Explanation:
(C) : Let, coordinate of \(\mathrm{P}\) be \((\mathrm{x}, \mathrm{y})\) \((\mathrm{BP})^{2}-(\mathrm{AP})^{2}=121\) \(\Rightarrow\left[(5-\mathrm{x})^{2}+(11-\mathrm{y})^{2}\right]-\left[(2-\mathrm{x})^{2}+(5-\mathrm{y})^{2}\right]=121\) \(\Rightarrow 25+\mathrm{x}^{2}-10 \mathrm{x}+121+\mathrm{y}^{2}-22 \mathrm{y}-4-\mathrm{x}^{2}+4 \mathrm{x}-25-\) \(\mathrm{y}^{2}+10 \mathrm{y}=121\) \(=-10 x-22 y-4+4 x+10 y=0\) \(-6 x-12 y-4=0\) \(-2(3 x+6 y+2)=0\) \(3 x+6 y+2=0\) Slope of the line is \(=-\frac{\text { coefficient of } x}{\text { coefficient } x^{2}}=\frac{-3}{6}=\frac{-1}{2}\)
AP EAMCET-2022-05.07.2022
Co-Ordinate system
88383
A value of \(\mathrm{k}\) such that the straight lines \(\mathrm{y}-3 \mathrm{kx}\) \(+4=0\) and \((2 k-1) x-(8 k-1) y-6=0\) are perpendicular is
88384
If \(m_{1}, m_{2}\left(m_{1}>m_{2}\right)\) are the slopes of the lines which make an angle of \(30^{\circ}\) with the line joining the points \((1,2)\) and \((3,4)\), then \(\frac{m_{1}}{m_{2}}=\)
88385
In an isosceles right angled triangle, if the equation of the hypotenuse and its opposite vertex are \(3 x+4 y=4\) and \((2,2)\), then the slopes of the remaining two sides are.
1 \(\frac{1}{7},-7\)
2 \(\frac{-1}{7}, 7\)
3 \(\frac{1}{7}, 7\)
4 \(\frac{-1}{7},-7\)
Explanation:
(A): Hypotenuse is along the line \(3 x+4 y-4=0\) \(\therefore \quad\) Slope of \(\mathrm{AC}=\frac{-3}{4}\) \(\because \quad \Delta \mathrm{ABC}\) is isosceles right angled triangle, Now, let the slope of the line making an angle \(45^{\circ}\) with \(\mathrm{AC}\) be \(\mathrm{m}-\) \(\tan 45^{\circ}=\left|\frac{m-\left(\frac{-3}{4}\right)}{1+m\left(\frac{-3}{4}\right)}\right|\) \(\therefore\) \(\pm 1=\frac{4 m+3}{4-3 m}\) \(4 m+3=4-3 m \quad \text { or } \quad 4 m+3=-4+3 m\) \(7 m=1\) \(m=\frac{1}{7}\) So, slope of remaining two sides \(\mathrm{BC}\) and \(\mathrm{AB}\) are \(\frac{1}{7}\) and -7 .
88382
Suppose a point \(P\) moves so that \(\mathbf{B P}^{2}-\mathbf{A P}^{2}=\) 121 , where \(A\) and \(B\) are \((2,5)\) and \((5,11)\) respectively. Then the locus of \(P\) is a straight line, whose slope is
1 \(1 / 2\)
2 -2
3 \(-1 / 2\)
4 2
Explanation:
(C) : Let, coordinate of \(\mathrm{P}\) be \((\mathrm{x}, \mathrm{y})\) \((\mathrm{BP})^{2}-(\mathrm{AP})^{2}=121\) \(\Rightarrow\left[(5-\mathrm{x})^{2}+(11-\mathrm{y})^{2}\right]-\left[(2-\mathrm{x})^{2}+(5-\mathrm{y})^{2}\right]=121\) \(\Rightarrow 25+\mathrm{x}^{2}-10 \mathrm{x}+121+\mathrm{y}^{2}-22 \mathrm{y}-4-\mathrm{x}^{2}+4 \mathrm{x}-25-\) \(\mathrm{y}^{2}+10 \mathrm{y}=121\) \(=-10 x-22 y-4+4 x+10 y=0\) \(-6 x-12 y-4=0\) \(-2(3 x+6 y+2)=0\) \(3 x+6 y+2=0\) Slope of the line is \(=-\frac{\text { coefficient of } x}{\text { coefficient } x^{2}}=\frac{-3}{6}=\frac{-1}{2}\)
AP EAMCET-2022-05.07.2022
Co-Ordinate system
88383
A value of \(\mathrm{k}\) such that the straight lines \(\mathrm{y}-3 \mathrm{kx}\) \(+4=0\) and \((2 k-1) x-(8 k-1) y-6=0\) are perpendicular is
88384
If \(m_{1}, m_{2}\left(m_{1}>m_{2}\right)\) are the slopes of the lines which make an angle of \(30^{\circ}\) with the line joining the points \((1,2)\) and \((3,4)\), then \(\frac{m_{1}}{m_{2}}=\)
88385
In an isosceles right angled triangle, if the equation of the hypotenuse and its opposite vertex are \(3 x+4 y=4\) and \((2,2)\), then the slopes of the remaining two sides are.
1 \(\frac{1}{7},-7\)
2 \(\frac{-1}{7}, 7\)
3 \(\frac{1}{7}, 7\)
4 \(\frac{-1}{7},-7\)
Explanation:
(A): Hypotenuse is along the line \(3 x+4 y-4=0\) \(\therefore \quad\) Slope of \(\mathrm{AC}=\frac{-3}{4}\) \(\because \quad \Delta \mathrm{ABC}\) is isosceles right angled triangle, Now, let the slope of the line making an angle \(45^{\circ}\) with \(\mathrm{AC}\) be \(\mathrm{m}-\) \(\tan 45^{\circ}=\left|\frac{m-\left(\frac{-3}{4}\right)}{1+m\left(\frac{-3}{4}\right)}\right|\) \(\therefore\) \(\pm 1=\frac{4 m+3}{4-3 m}\) \(4 m+3=4-3 m \quad \text { or } \quad 4 m+3=-4+3 m\) \(7 m=1\) \(m=\frac{1}{7}\) So, slope of remaining two sides \(\mathrm{BC}\) and \(\mathrm{AB}\) are \(\frac{1}{7}\) and -7 .
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Co-Ordinate system
88382
Suppose a point \(P\) moves so that \(\mathbf{B P}^{2}-\mathbf{A P}^{2}=\) 121 , where \(A\) and \(B\) are \((2,5)\) and \((5,11)\) respectively. Then the locus of \(P\) is a straight line, whose slope is
1 \(1 / 2\)
2 -2
3 \(-1 / 2\)
4 2
Explanation:
(C) : Let, coordinate of \(\mathrm{P}\) be \((\mathrm{x}, \mathrm{y})\) \((\mathrm{BP})^{2}-(\mathrm{AP})^{2}=121\) \(\Rightarrow\left[(5-\mathrm{x})^{2}+(11-\mathrm{y})^{2}\right]-\left[(2-\mathrm{x})^{2}+(5-\mathrm{y})^{2}\right]=121\) \(\Rightarrow 25+\mathrm{x}^{2}-10 \mathrm{x}+121+\mathrm{y}^{2}-22 \mathrm{y}-4-\mathrm{x}^{2}+4 \mathrm{x}-25-\) \(\mathrm{y}^{2}+10 \mathrm{y}=121\) \(=-10 x-22 y-4+4 x+10 y=0\) \(-6 x-12 y-4=0\) \(-2(3 x+6 y+2)=0\) \(3 x+6 y+2=0\) Slope of the line is \(=-\frac{\text { coefficient of } x}{\text { coefficient } x^{2}}=\frac{-3}{6}=\frac{-1}{2}\)
AP EAMCET-2022-05.07.2022
Co-Ordinate system
88383
A value of \(\mathrm{k}\) such that the straight lines \(\mathrm{y}-3 \mathrm{kx}\) \(+4=0\) and \((2 k-1) x-(8 k-1) y-6=0\) are perpendicular is
88384
If \(m_{1}, m_{2}\left(m_{1}>m_{2}\right)\) are the slopes of the lines which make an angle of \(30^{\circ}\) with the line joining the points \((1,2)\) and \((3,4)\), then \(\frac{m_{1}}{m_{2}}=\)
88385
In an isosceles right angled triangle, if the equation of the hypotenuse and its opposite vertex are \(3 x+4 y=4\) and \((2,2)\), then the slopes of the remaining two sides are.
1 \(\frac{1}{7},-7\)
2 \(\frac{-1}{7}, 7\)
3 \(\frac{1}{7}, 7\)
4 \(\frac{-1}{7},-7\)
Explanation:
(A): Hypotenuse is along the line \(3 x+4 y-4=0\) \(\therefore \quad\) Slope of \(\mathrm{AC}=\frac{-3}{4}\) \(\because \quad \Delta \mathrm{ABC}\) is isosceles right angled triangle, Now, let the slope of the line making an angle \(45^{\circ}\) with \(\mathrm{AC}\) be \(\mathrm{m}-\) \(\tan 45^{\circ}=\left|\frac{m-\left(\frac{-3}{4}\right)}{1+m\left(\frac{-3}{4}\right)}\right|\) \(\therefore\) \(\pm 1=\frac{4 m+3}{4-3 m}\) \(4 m+3=4-3 m \quad \text { or } \quad 4 m+3=-4+3 m\) \(7 m=1\) \(m=\frac{1}{7}\) So, slope of remaining two sides \(\mathrm{BC}\) and \(\mathrm{AB}\) are \(\frac{1}{7}\) and -7 .
